InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1951. |
If P(A)shows the probability of occuring an event, then :(A) P(A) < 0 (B) P(A) > 1(C) 0 ≤ P(A) ≤ 1(D) -1 ≤ P(A) ≤ 1 |
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Answer» Answer is (C) 0 ≤ P(A) ≤ 1 |
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| 1952. |
On expressing the probability of occuring an event in the terms of percentage. It can never be taken place.(A) less than 100(B) less than 0(C) more than 1(D) Besides a whole number everything |
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Answer» Answer is (B) less than 0 |
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| 1953. |
A can solve 90% of the problems given in a book and B can solve 70%.What is the probability that at least one of them will solve theproblem, selected at random from the book? |
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Answer» Let `E_(1)=` event that A solves the problem, and `E_(2)=` event that B solves the problem. Then, `P(E_(1))=90/100=9/10` and `P(E_(2))=70/100=7/10`. Clearly, `E_(1)` and `E_(2)` are independent events. `:. P(E_(1) nn E_(2))=P(E_(1))xxP(E_(2))=(9/10xx7/10)=63/100`. `:.` P(at least one of them will solve the problem) `=p(E_(1) uu E_(2))` `=P(E_(1))+P_(E_(2))-P(E_(1) nn E_(2))` `=(9/10+7/10-63/100)=((90+70-63))/100=97/100`. Hence, the required probability is 0.97. |
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| 1954. |
Prove that inthrowing a pair of dice, the occurrence of the number 4 on the first die isindependent of the occurrence of 5 on the second die. |
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Answer» Here, `4` will occur on first die. But, it will not impact any of the occurance of second die. Second die can contain any number between `1` and `6` including `5`. As, number occuring on first die do not have any impact on number occuring `5`, Therefore, the occurrence of the number `4` on the first die is independent of the occurrence of `5` on the second die. |
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| 1955. |
In a school there are 1000 students, out of which 430 are girls. It isknown that out of 430, 10% of the girls study in class XII. What is theprobability that a student chosen randomly studies in class XII given thatthe chosen student is a girl? |
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Answer» Here, `A = `Event that chosen student is a girl. `:. P(A) = 430/1000 = 43/100` Let `B = `Event that chosen student studies in class XII. `P(B) = 10% = 10/100 = 1/10` Now, `P(AnnB) =` Probaility chosen student is from class XII and is a girl. `=>P(AnnB) = 43/100**1/10 = 43/1000` `:.` Required probability , `P(B/A) = (P(AnnB) )/(P(A)) = 43/1000**100/43 = 1/10.` |
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| 1956. |
In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl? |
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Answer» here we have to find `P(A|B) = ?` Given that `P(A)=430/1000` `P(A nn B) = ((10/100)430)/1000 ` `=43/1000` `P(B|A) = (P( A nn B)) /(P(A))` `=43/1000*1000/430 = 1/10` answer |
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| 1957. |
One mapping is selected at random from all the mappings of the set A = {1, 2, 3, ….., n) into itself. The probability that the mapping selected is one to one isA. \(\frac{1}{n^2}\)B. \(\frac{1}{n!}\)C. \(\frac{(n-1)!}{n^{n-1}}\) D. none of these |
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Answer» We know that the no of mapping from a set A to same set containing n elements = nn As each element have n options to be mapped with n elements. For mapping selected to be one to one, for first element we have n options ,for second we have n-1 options an so on… ∴ total such mappings = n × (n-1)× (n-2)× … = n! ∴ P(mapping is one to one) = \(\frac{n!}{n^n}\) = \(\frac{n(n-1)!}{n^n}\) = \(\frac{(n-1)!}{n^n-1}\) Our answer matches with option (c) ∴ Option (c) is the only correct choice |
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| 1958. |
The probability of getting a sum of atleast 9 in one throw of two dice is:A. `1/3`B. `11/36`C. `5/18`D. `13/18` |
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Answer» Correct Answer - C N/a |
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| 1959. |
In a hurdle race, a runner has probability `p` of jumping over a specific hurdle. Given that in `5` trials, the runner succeeded `3` times, the conditional probabilit that the runner had succeeded in the first trial isA. `3//5`B. `2//5`C. `1//5`D. None of these |
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Answer» Correct Answer - A `(a)` Let `A` denote the event that the runner succeeds exactly `3` times out of five and `B` denote the events that the runner suceeds on the first trial. `P(B//A)=(P(BnnA))/(P(A))` But `P(BnnA)=P` (clearing succeeding in the first trial and exactly once in two other trials) `=^(4)C_(2)p^(2)(1-p)^(2))=6p^(3)(1-p)^(2)` and `P(A)=^(5)C_(3)p^(3)(1-p)^(2)=10p^(3)(1-p)^(2)` Thus, `P(B//A)=(6p^(3)(1-p)^(2))/(10p^(3)(1-p)^(2))=(3)/(5)` |
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| 1960. |
A random variable X is defined by `X={{:("3 with probability"=1/3),("4 with probability"=1/4),("12 with probability"=5/(12)"):}` Then, E(X) isA. 6B. 7C. 5D. 8 |
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Answer» Correct Answer - B `E(X)=3xx1/3+4xx1/4+12xx5/(12)=7` |
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| 1961. |
Let the random variable X is defined as time (in minutes) that elapses between the bell and end of the lecture in case of collagen professor whrer pdf is defined as `f(x)={{:(kx^2","0lexlt2),(0", ""elsewhere"):}` find the probability that lecture continue for atleast 90s beyond the bellA. `(37)/(64)`B. `(35)/(64)`C. `(33)/(69)`D. None of these |
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Answer» Correct Answer - A We known that `oversetoounderset(-oo)intf(x)dx=1` `therefore" "0+underset0overset2intf(x)dx=1` `rArr" "underset0overset2intkx^2dx=1` `rArr" "k[x^3/3]_0^2=1rArr[8/3]=1` `rArr" "k=3/8` Clearly, the probability that the lecture continuous for at least 90s i.e.`3/2` min beyond the bell `=P(3/2lexle2)=overset2underset(3//2)intf(x)dx=k""overset2underset(3//2)intx^2dx` `=k[x^3/3]_(3//2)^2=k/3[8-(27)/8]=(37k)/(24)` `=(37xx3/8)/(24)=(37)/(64)` |
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| 1962. |
Let X is a continuous random variable with probability density function `f(x)={{:(x/6+k,0lexle3),(0," otherwise"):}` The value of k is equal toA. `1/(12)`B. `1/3`C. `1/4`D. `1/6` |
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Answer» Correct Answer - A `oversetoounderset(-oo)intf(x)dx=1rArr0+overset3underset0int(x/6+k)dx+0=1` `rArr" "[x^2/(12)+kx]_0^3=1rArr3/4+3k=1` `rArr" "3k=1/4rArrk=1/(12)` |
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| 1963. |
If the probability density function of a random variable X is given as then F(0) is equal to A. `P(Xlt0)`B. `P(Xgt0)`C. `1-P(Xgt0)`D. `1-PXlt0)` |
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Answer» Correct Answer - C We know that, cumulative distribution function `F(x)=P(Xlex)` `therefore" "F(0)=P(Xle0)` `=P(x=0)+P(x=-1)+P(x=-2)` `=0.15+0.3+0.2=0.65` (a) `P(Xlt0)=P(x=-1)+P(x=-2)` `=0.3+0.2=0.5` `therefore" "P(Xlt0)neF(0)` (b) `P(Xgt0)=P(x=1)+P(x=2)` `0.25+0.1=0.65` `therefore" "P(Xgt0)neF(0)` (c ) `1-P(Xgt0)=1-0.35=0.65` `therefore" "1-P(Xgt0)=F(0)` Hence, option (c ) is correct |
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| 1964. |
The life in hours of a ratio tube is continuous random variable with pdf `f(x)={{:((100)/x^2","xge100),(0",""else where"):}` Then, the probability that the life of tube will than 200 h if it is known that th tube is still functioning after 150 h of services isA. `1/4`B. `1/3`C. `1/2`D. None of these |
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Answer» Correct Answer - A `thereforeP[(xlt200)|(xgt150)]` `=(P(150ltxlt200))/(P(xgt150))` `(overset(200)underset(150)int(100)/x^2dx)/(oversetoounderset(150)int(100)/x^2dx)=([(-100)/x]_(150)^(200))/([(-100)/x]_(150)^oo)` `=(-[(100)/(200)-(100)/(150)])/(-[0-(100)/(150)])=(1/6)/(2/3)=1/4` |
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| 1965. |
Let A and B be two independent events. Statement-1: If P(A)=0.3 and `P(A cup overline(B))=0.8`, then `P(B)=(2)/(7)` Statement-2: For any event `E, P(overline(E ))=1-P(E )`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A Clearly, statement-2 is true. If A and B are independent events such that `P(A cup overline(B))=0.8` `implies 1-P(overline(A))P(overline(B))=0.8` `implies 1-(1-P(A))P(B)=0.8` `implies 1-0.7P(B)=0.8 implies P(B)=(2)/(7)` So, statement-1 is true and statement-2 is a correct explanation for statement-1. |
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| 1966. |
The digits 1,2,3,4,5,6,7,8 and 9 are written in random order to form a nine digit number. The probability that this number is divisible by 4, isA. `(1)/(9)`B. `(2)/(3)`C. `(2)/(9)`D. `(7)/(9)` |
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Answer» Correct Answer - C There are `.^(9)P_(9)=9!` ways of arranging the given digits to form a nine digit number. So, total number of 9 digit numbers formed by the given digits =9!. Out of these 9! Numbers only those numbers are divisble by 4 which have their last digits as even natural number and the numbers formed by their last two digits are divisible by 4. The various possibilities of last two digits are : 12, 32, 52, 72, 92 24, 64, 84 16, 36, 56, 76, 96 28, 48, 68 This means that there are 16 ways of choosing the last two digits. Corresponding to each of these ways the remaining 7 digits can be arranged in 7! ways. Therefore, the total number of 9 digit numbers divisble by 4 is `16xx7!` Hence, required probability `=(16xx7!)/(9!)=(2)/(9)` |
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| 1967. |
एक विशेष समस्या को A और B द्वारा स्वतंत्र रूप से हल करने की प्रायिकताएं क्रमशः `1/2` और `1/3` हैं। यदि दोनों स्वतंत्र रूप से समस्या हल करने का प्रयास करते हैं। तो प्रायिकता ज्ञात कीजिए कि (i) समस्या हल हो जाती है (ii) उनमें से तथ्यतः कोई एक समस्या हल कर लेता है।A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A Clearly, statement-2 is true. Required probability `=1-(1-(1)/(2))(1-(1)/(3))(1-(1)/(4))=(3)/(4)` So, statement-1 is also true and statement-2 is a correct explanation for statement-1. |
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| 1968. |
Two coins are tossed together. Find the probability of getting both heads or both tails. |
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Answer» Possibilities are HH, HT, TH, TT P(HH or TT) = 2/4 = 1/2 |
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| 1969. |
Three dice are thrown together. Find the probability of getting a totalof at least 6. |
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Answer» `6*6*6=216` `P_1=1-P` `{(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,2,2),(2,2,1),(2,1,2),(1,1,3),(3,1,1),(1,3,1)}` P(sum<6)=10/216 P(sum>6)=1-P(sum <6) `=1-10/216=206/216=103/108`. |
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| 1970. |
If I toss a coin 3 times and get head each time, then I should expect a tail to have higher chance in the 4th toss.State true or false and give the reason |
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Answer» False, because of the outcomes 'head' and 'tail' are equally, likely. So, every time the probability of getting head or tail is 1/2. |
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| 1971. |
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house , isA. `8/9`B. `7/9`C. `2/9`D. `1/9` |
| Answer» Correct Answer - D | |
| 1972. |
A man draws a card from a pack of 52 cards and then replaces it. After shuffling the pack, he again draws a card. This he repeats a number of times. The probability that he will draw a heart for the first time in the third draw isA. `(9)/(64)`B. `(27)/(64)`C. `(1)/(4)xx(.^(39)C_(2))/(.^(52)C_(2))`D. none of these |
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Answer» Correct Answer - A Let `A_(i)` be the event of drawing a heart card in ith draw. Then, `P(A_(i))=(.^(13)C_(1))/(.^(52)C_(1))=(1)/(4), i=1,2,3` Required probability `=P(overline(A_(1)) cap overline(A_(2)) cap A_(3))` `=P(overline(A_(1)))P(overline(A_(2)))P(A_(3))=(3)/(4)xx(3)/(4)xx(1)/(4)=(9)/(64)` |
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| 1973. |
A person draws a card from a well shuffled pack of 52 playing cards. Replaces it and shuffles the pack. He continues doing so until he draws as pade. The chance that he fails first two times isA. `(1)/(16)`B. `(9)/(16)`C. `(9)/(64)`D. `(1)/(64)` |
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Answer» Correct Answer - B Let `A_(i)` be the event that the person fails in `i^(th)` draw. Then, `P(A_(i))=1-(13)/(52)=(3)/(4), i=1,2,3,..` `therefore` Required Probability `=P(A_(1) cap A_(2))=P(A_(1))P(A_(2))` `implies` Required probability `=(3)/(4)xx(3)/(4)=(9)/(16)` |
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| 1974. |
Sid draws a card from a pack of cards, replaces it, shuffles the pack and then draws another card. What is the probability that the cards are both aces? |
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Answer» P(ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\) (\(\because\) there are 4 aces in a pack of 52 cards) \(\therefore\) P(both aces) = P(first ace and second ace) = \(\frac{1}{13}\) x \(\frac{1}{13}\)= \(\frac{1}{169}\) |
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| 1975. |
Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail. |
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Answer» In the experiment of tossing a coin once, the number of possible outcomes is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of outcomes favorable to E, (i.e., of getting a head) is 1. Therefore, P(E) = P (head) =Number of outcomes favourable to E/Number of all possible outcomes =1/2 Similarly, if F is the event ‘getting a tail’, then P(F) = P(tail) =1/2 |
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| 1976. |
Three houses are available in a locality. Three persons apply for thehouses. Each applies for one houses without consulting others. Theprobability that all three apply for the same houses isA. `1//9`B. `2//9`C. `7//9`D. `8//9` |
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Answer» Correct Answer - A Required probability = `("No. of favorable cases")/("Total no. of exhaustive cases")` = `(3)/(3xx3xx3) = (1)/(9)` |
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| 1977. |
A draws a card from a pack of `n`cards marked `1,2,ddot,ndot`The card is replaced in the pack and `B`draws a card. Then the probability that `A`draws a higher card than `B`is`(n+1)2n`b. `1//2`c. `(n-1)2n`d. none of theseA. `(n + 1)//2n`B. `1//2`C. `(n - 1)//2n`D. none of these |
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Answer» Correct Answer - C If A draws card higher than B, then the number of favorable cases is (n - 1) + (n + 2) + … + 3 + 2 + 1 (as when B draws card number 1, then A can draw any card from 2 to n and so on). Therefore, the required probability is `(n((n-1))/(2))/(n^(2)) = (n-1)/(2n)` |
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| 1978. |
Find the probability of getting a head when a coin is tossed once.Also find the probability of getting a tail. |
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Answer» When a coin is tossed once, the sample space is given by `S = {H, T}.` Let E be the event of getting a head. Then, `E = {H}.` `therefore n(E) = 1 and n(S) = 2` `rArr` P (getting a head) `= P(E) = (n(E))/(n(S)) = 1/2.` |
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| 1979. |
A coin is tossed once. Find the probability of getting a tail. |
| Answer» Correct Answer - `1/2` | |
| 1980. |
A coin is tossed once. Find the probability of getting a tail. |
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Answer» We know that Probability of occurrence of an event = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) Total outcomes of the coin are tails and heads Hence the total no. of outcomes are 2 (i.e. heads and tails) And the desired output is tail. Hence no. of desired outcomes = 1 Therefore, the probability of getting a tail is = \(\frac{1}{2}\) Conclusion: Probability of getting a tail when a coin is flipped is 0.5 or \(\frac{1}{2}\) |
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| 1981. |
A coin is tossed once. What is the probaility of getting a head ? |
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Answer» When a coin is tossed once, all possible outcomes are H and T. Total number of possible outcomes = 2. The favourable outcome is H. Number of favourable outcomes = 1. `:. ` P(getting a head) = P(H) = `("number of favourable outcomes")/("total number of possible outcomes ") = 1/2`. |
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| 1982. |
The probability of “the last day of a month is Sunday” A) 2/7B) 1/7C) 3/7D) 4/7 |
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Answer» Correct option is (B) 1/7 Last day of a month can be any of 'Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday'. \(\therefore\) Total outcomes is n(S) = 7 The favourable outcomes for the event that "the last day of a month is Sunday" is only Sunday. \(\therefore\) Number of favourable outcome is n(E) = 1 \(\therefore\) The probability of "the last day of a month is Sunday" is \(P(E)=\frac{n(E)}{n(S)}=\frac17.\) Correct option is B) 1/7 |
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| 1983. |
The probability that in a family of 5 members,exactly two members have birthday on sunday is:-A. `(12 xx 5^(3))//7^(5)`B. `(10 xx 6^(2))//7^(5)`C. `2//5`D. `(10 xx 6^(3))//7^(5)` |
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Answer» Correct Answer - D A person can have his/her birthday on any one of the seven days of the week. So 5 persons can have their birthdays in `7^(5)` ways. Out of 5, three persons can have their birthdays on days other than Sundays in `6^(3)` ways and other 2 on Sundays. Hence, the required probability is `=(.^(5)C_(2) xx 6^(3))/(7^(5)) = (10 xx 6^(3))/(7^(5))` (Note that 2 persons can be selected out of 5 in `.^(5)C_(2)` ways). |
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| 1984. |
A coin is tossed once. What is the probaility of getting a tail? |
| Answer» Correct Answer - ` 1/2` | |
| 1985. |
From the letters of the word “MOBILE”, if a letter is selected, what is the probability of getting the letter vowel ?A) 1/7B) 1/3C) 1/2D) 1/4 |
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Answer» Correct option is: C) \(\frac{1}{2}\) Total letters in the word "MOBILE"= 6, Total vowel letters in the word "MOBILE" = 3 (\(\because\) O, I & E are vowels) \(\therefore\) Probability of getting the vowel letter = \(\frac {Total \,vowel \,letters}{Total\, No \,of \,letters}\) = \(\frac 36 = \frac 12\) Correct option is: C) \(\frac{1}{2}\) |
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| 1986. |
A baby is born the probability that it is a boy or girl isA) 1B) 1/2C) −1/2D) 1/3 |
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Answer» Correct option is: B)\(\frac{1}{2}\) The probability that born baby is a boy = The probability that born baby is a girl = \(\frac{1}{2}\) Correct option is: B) \(\frac{1}{2}\) |
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| 1987. |
Fill in the blanks.(i) Probability of an impossible event = …….(ii) Probability of a sure event = ……..(iii) Let E be an event. Then, P (not E) = ……..(iv) P (E) + P (not E) = ……..(v) …….. ≤ P (E) ≤ ………. |
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Answer» (i) Probability of an impossible event = 0. (ii) Probability of a sure event = 1. (iii) Let E be an event. Then, P (not E) = 1 – P (E). (iv) P (E) + P (not E) = 1. (v) 0 ≤ P (E) ≤ 1. |
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| 1988. |
_____ is the probability of an even that in any year, Sunday is on 15 August? |
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Answer» 15th August may fall in any days of the week. Total outcomes = 7 If Sunday comes on 15th August, then possible outcomes = 1 P(E) = possible outcomes / Total outcomes P(E) = 1/7 |
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| 1989. |
Fill in the blanks: (i) Probability of a sure event is ………. . (ii) Probability of an impossible event is ………….. . (iii) The probability of an event (other than sure and impossible event) lies between ……….. . (iv) Every elementary event associated to a random experiment has ………. Probability. (v) Probability of an event A+ Probability of event ‘not A’ = ………….. .(vi) Sum of the probabilities of each outcome in an experiment is …………. . |
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Answer» (i) 1 (ii) 0 (iii) 0 and 1 (iv) Equal (v) 1 (vi) 1 |
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| 1990. |
Fill in the blanks . (i) Probability of an impossible event= …….. (ii) Probability of a sure event = …… (iii) Let E be the event . Then , P(not E) = ……. (iv) P(E) + P(not E) = ……. . (v) ……….`le P(E) le ` ……. . |
| Answer» Correct Answer - (i) 0 (ii) 1 (iii) 1-P(E) (iv) 1 (v) 0, 1 | |
| 1991. |
Gita said that the probability of impossible event is 1. Pravallika said that probability of sure event is 0 and Atiya said that the probability of any event lies between 0 arid 1. In the above with whom you will agree ?A) Gita B) Pravallika C) Atiya D) All the three |
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Answer» Correct option is: C) Atiya \(\because\) Probability of an impossible event is 0 \(\neq\) 1. Hence, the statement of Gita is wrong. Also, probability of a sure event is 1 \(\neq\) 0 Hence, the statement of Pravallika is wrong. Since, 0 \(\leq\) P \(\leq\) 1, i.e, probability of any event always lies between 0 and 1. Hence, the statement of Atiya is true Thus, we have to agree with Atiya. Correct option is: C) Atiya |
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| 1992. |
Fill in the blacks: (i) The probability of an impossible event is ………. . (ii) The probability of a sure event is …….. . (iii) For any event E, P(E ) + P(not E)=……… . (iv) The probability of a possible but not a sure event lies between .......... and ......... . (v) The sum of probabilities of all the outcomes of an experiment is ......... . |
| Answer» Correct Answer - (i) 0 (ii)1(iii)1(iv) 0,1 (v) 1 | |
| 1993. |
In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:(i) winning of Mahesh (ii) winning of John |
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Answer» (i) But if John looses, Mahesh wins Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only Therefore, P(winning of Mahesh) = 0.54 (ii) P(winning of Mahesh) + P(winning of John) = 1 0.54 + P(winning of John) = 1 P(winning of John) = 1 – 0.54 P(winning of John) = 0.46 |
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| 1994. |
What is the probability of drawing a King or a Queen or a Jack from a deck of cards? |
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Answer» Number of cards n(S) = 52 No. of King cards n(A) = 4 No. of Queen cards n(B) = 4 No. of Jack cards n(C) = 4 Probability of drawing a King card = \(\frac{n(A)}{n(S)}=\frac{4}{52}\) Probability of drawing a Queen card = \(\frac{n(B)}{n(S)}=\frac{4}{52}\) Probability of drawing a Jack card = \(\frac{n(C)}{n(S)}=\frac{4}{52}\) The Probability of drawing a King or a Queen or a Jack from a deck of cards = P(A) + P(B) + P(C) = \(\frac{4}{52}+\frac{4}{52}+\frac{4}{52}=\frac{4+4+4}{52}=\frac{12}{52}=\frac{3}{13}\) |
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| 1995. |
Two integers are selected at random from the set {1,2,…………..,11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even isA. `(2)/(5)`B. `(1)/(2)`C. `(7)/(10)`D. `(3)/(5)` |
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Answer» Correct Answer - A In {1,2,3,……,11} there are 5 even numbers and 6 odd numbers . The sum even is possible only when both are odd or both are even . Let A be the event that denotes both numbers are even and B be the event that denotes sum of numbers is even . Then , `n(A)=""^(5)C_(2)+""^(6)C_(2)` . Required probability `P(A//B)=(P(A cap B))/(P(B)) =(""^(5)C_(2)//""^(11)C_(2))/(((""^(6)C_(2)+""^(5)C_(2)))/(""^(11)C_(2)))=(""^(5)C_(2))/(""^(6)C_(2)+""^(5)C_(2))=(10)/(15+10)=(2)/(5)` |
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| 1996. |
Let `S={1,2,...,20}` A subset `B` of S is said to be `"nice"`, if the sum of the elements of `B` is 203. Then the probability that a randomly chosen subset of `S` is `"nice"` is: (a) `7/(2^20)` (b) `5/(2^20)` (c) `4/(2^20)` (d) `6/(2^20)`A. `(6)/(2^(20))`B. `(4)/(2^(20))`C. `(7)/(2^(20))`D. `(5)/(2^(20))` |
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Answer» Correct Answer - D Number of subset of `S=2^(20)` Sum of elements in S is `1+2+......+20=(20(21))/2=210 ` `[therefore 1+2+......+n=(n(n+1))/2]` Clearly , the sum of elements of a sunset would be 203 , if we consider it as follows S -{7},S --·{1,G}S-{2,5},S-{3,4} S-{1,2,4) `therefore` Number of a favoueables cases =5 Hence , required probility `=5/(2^(20)` |
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| 1997. |
A bag contains `n+1`coins. If is known that one of these coins shows heads onboth sides, whereas the other coins are fair. One coin is selected at randomand tossed. If the probability that toss results in heads is 7/12, then findthe value of `ndot`A. 3B. 4C. 5D. none of these |
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Answer» Correct Answer - C Let `E_(1)` denote the event a coin with two heads is selected and `E_(2)` denote the event a fair coin is selected . Let A be the event the toss results in head. Then., `P(E_(1))=(1)/(n+1),P(E_(2))=(n)/(n+1),P(A//E_(2))=1` and `P(A//E_(2))=1//2`. `therefore P(A)=P(E_(1))P(A//E_(1))+P(E_(2))P(A//E_(2))` `implies (7)/(12)=(1)/(n+1)xx1+(n)/(n+1)xx(1)/(2)` `implies 12+6n=7n+7 implies n=5` |
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| 1998. |
A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is selected at random from the bag and tossed. The probability of getting a head isA. `9//16`B. `11//16`C. `5//9`D. none of these |
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Answer» Correct Answer - A Let A be the event of selecting a counterfeit coin and B be the event of getting head. Then, Required probability `=P(A cap B) cup (overline(A) cap B)` `=P(A cap B)+P(overline(A))P(B//overline(A))` `=(2)/(16)xx1+(14)/(16)xx(1)/(2)=(9)/(16)` |
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| 1999. |
A box contains 10 mangoes out of which 4 arerotten. Two mangoes are taken out together. If one of them is found to begood, then find the probability that the other is also good.A. `1//3`B. `8//15`C. `5//13`D. `2//3` |
| Answer» Correct Answer - C | |
| 2000. |
A lot consists of 12 good pencils, 6 with minor defects and 2 with major defects. A pencil is choosen at random. The probability that this pencil is not defective, isA. `3//5`B. `3//10`C. `4//5`D. `1//2` |
| Answer» Correct Answer - A | |