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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.
| 1. |
एक प्रयोग के सफल होने का संयोग उसके असफल होने से दो गुना है। प्रायिकता ज्ञात कीजिए कि अगले छः परीक्षणों में कम से कम 4 सफल होंगे।A. `(192)/(729)`B. `(256)/(729)`C. `(240)/(729)`D. `(4962)/(729)` |
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Answer» Correct Answer - B Let p be the probability of success and q that of failure. It given that . `p=2qrArr 2q=1-qrArr q=(1)/(3) and =(2)/(3)` Let X denote the number of successes in 6 trails. Then, ltbgt `P(X=r)=.^6C_(r)p^rq^6-r=.^6C_(r)((2)/(3))^r((1)/(3))^(6-r), r=0,2,.......6` Required Probability `=P(Xge 5)=P(X=5)+P(X=6)` `.^6C_(5)((2)/(3))^5((1)/(3))^(6-5)+.^6C_(6)((2)/(3))^6((1)/(3))^(0)=(256)/(729)`. |
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| 2. |
The probability that an event A happens in one trial of an experiment, is 0.4 There independent trials of the experiments are performed. The probability that the event A happens atleast once, isA. `0.936`B. `0.784`C. `0.904`D. none of these |
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Answer» Correct Answer - B Required Probability `=1-P` (The event does not happen in any trails) `=1-(0.6)^3=0.784` . |
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| 3. |
The mean and variance of a random variable X having a binomial distribution are `4 and 2` respectively. The `P(X=1)` isA. `1//4`B. `1//32`C. `1//16`D. `1//8` |
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Answer» Correct Answer - B Let n and p be the parameters of the binomial distribution. We have, Mean `=4` and Variance `=2` `rArr np=4 and npq=2` `rArr p=1//2 =q and n=8` `therefore` Required Probability `=P(X-1)=.^8C_(1)((1)/(2))^1((1)/(2))^7=(1)/(32)` |
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| 4. |
India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points `0, 1 and 2` are `0.45, 0.05 and 0.50` respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is (a) `0.8750` (b) `0.0875` (c) `0.0625` (d) `0.0250`A. `(1)/(80)`B. `(7)/(80)`C. `(7)/(8)`D. `(1)/(8)` |
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Answer» Correct Answer - B Since there are four matches to be played. So, India can get a maximum of 8 points. `therefore` P(Inida gets atleast 7 points) `=P` (Getting exactly 7 points) `+P` (Getting exactly 8 points) `=P` (Getting 2 points in each of the three matches and 1 in one match) `+P` (Getting 2 points in each of the four matches) `=.^4C_(3)(0.5)^3(0.05)^1+.^4C_(4)(0.5)^4` `=4xx(0.5)^3xx(0.05)+(0.5)^4=(0.5)^3(0.2+0.5)=(7)/(80)` |
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| 5. |
The value of C for which `P(X=k)=Ck^2` can serve as the probability function of a random varibale X that takes value `0,1,2,3,4,`isA. `(1)/(30)`B. `(1)/(10)`C. `(1)/(3)`D. `(1)/(15)` |
| Answer» Correct Answer - A | |
| 6. |
A man takes a step forward with probability `0.4` and backward with probability `0.6`. The probability that at the end of eleven steps he is just one step away from the starting point, isA. `.^11(C)_(5)(0.4)^6(0.6)^5`B. `.^11(C)_(6)(0.4)^5(0.6)^6`C. `.^11(C)_(5)(0.4)^5(0.6)^6`D. `.^n(C)_(5)(0.4)^5(0.6)^5` |
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Answer» Correct Answer - C Let p denote the probability that the mant eks a step foraward. Then, `p=0.4` `therefore q=1 -p=1-0.4=0.6` Let X denote the number of steps taken in the forward direction. Since the stpes are idependent of each other, therefore X is a binomial variate with parameters `n=11` and `p=0.4` such that `P(X=r)=.^11C_r(0.4)^r(0.6)^11 -r, r=0,1,2,.....,11 ..(i)` Since the man is one step away from the initial poin, he is either one step forward or one step backward from the initial point at the end of eleven steps. If he is one step forward, then he must have taken six steps, forward and five steps backward and if he is one step backward, then he must have taken five steps forward and six backward. Thus , either `X=6 or X=5`. `therefore` Required Probability `=P[(X=5)or (X=6)]` `P(X=5)+P(X=6)` `=.^11C_5(0.4)^5(0.6)^(11-5)+.^11C_6(0.6)^6(0.6)^(11-6)["Using"(i)]` `=.^11C_5(0.4)^5(0.6)^5[0.6_0.4]` `[because .^11C_5=.^11C_6]` `=.^C_5(0.4)^5(0.6)^5` |
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| 7. |
A coin is tossed n times. The probability of getting head at least once is greater than `0.8.` Then the least value of n isA. 7B. 6C. 5D. 3 |
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Answer» Correct Answer - D Suppose the coin is tossed n times. Let X be the number of heads obtained. Then, X follows a binomial distribution with parameters n and `p=1//2`. Now, `P(Xge1)ge0.8` `rArr 1-P(X=0)ge0.8` `rArr 1-.^nC_(0)p^0(1-p)^nge 0.8` `rArr ((1)/(2))^nle 0.2rArr ((1)/(2))^nle (1)/(5)rArr 2^nge 5rArr nge 3` Hence, the least value of n is 3. |
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| 8. |
From a box containing 20 tickets marked with numbers 1 to 20, four tickets are drawn one by one. After each draw, the ticket is replaced. The probability that the largest value of tickets drawn is 15 is.A. `((3)/(4))^4`B. `(27)/(320)`C. `(27)/(1280)`D. none of these |
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Answer» Correct Answer - B We,have Probability of drawing a ticket bearing number `15 is (1)/(20)`. Probability of drawing a ticket bearing a number less than or equal to `15 is (15)/(20)=(3)/(4)`. `therefore` Required probability `=` Probability of drawing one ticket bearning number 15 and three tickets bearing numbers less than or equal to `15 =.^4C_(1)xx(1)/(20)xx((3)/(4))^3=(27)/(320)` |
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| 9. |
A coin is tossed 3n times. The chance that the number of times one gets head is not equal to the number of times one gets tail, isA. `.^(2n)C_(n)((1)/(2))^(2n)`B. `1-.^(2n)C_(n)`C. `1-.^(2n)C_(n)((1)/(4^n))`D. none of these |
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Answer» Correct Answer - C Required Probability `=1-` Probability of getting equal number of heads and tails `=1-` Probability of getting n heads and n tails `=1-.^2nC_(n)((1)/(2))^n((1)/(2))^(2n-n)=1-.^2nC_(n)((1)/(4^n))`. |
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| 10. |
A card is drawn from a pack of 52 playing cards. The card is replaced and the pack is reshuffled. If this is done six times. The probability that 2 hearts, 2 diamond and 2 black cards are drawn isA. `90xx((1)/(4))^6`B. `(45)/(2)xx((3)/(4))^4`C. `90xx((1)/(2))^10`D. none of these |
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Answer» Correct Answer - C We have, Probability of getting a heart in a draw `=(13)/(52)=(1)/(4)` Probability of getting a diamond in a draw `=(13)/(52)=(1)/(4)` Probability of getting a black card in a draw `=(26)/(52)=(1)/(2)` In 6 draws, 2 draws must contain hearts, 2 draws must contain diamond cards and 2 draws must contain black cards. This can happen in `.^6C_2xx.^4C_2xx .^C_2` mutually exculusive ways and the probability of each such way is `((1)/(4))^2xx((1)/(4))^2xx((1)/(4))^2` Hence, required probability `=.^6C_2xx.^4C_(2)xx((1)/(4))^2xx((1)/(4))^2xx((1)/(2))^2=90xx((1)/(2))^10` |
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| 11. |
If the range ot a random vaniabie `X` is `0,1,2,3,` at `P(X=K)=((K+1)/3^k)` a for `k>=0,` then a equalsA. `(2)/(3)`B. `(4)/(9)`C. `(8)/(27)`D. (16)/(81)` |
| Answer» Correct Answer - B | |
| 12. |
The number of times a die must be tossed to obtain a 6 at least one with probability exceeding `0.9` is at leastA. 13B. 19C. 25D. none of these |
| Answer» Correct Answer - A | |
| 13. |
The probability that a man can hit a target is `3//4`. He tries 5 times. The probability that he will hit the target at least three times isA. 291/364B. 371/464C. 471/502D. 459/512 |
| Answer» Correct Answer - D | |
| 14. |
If random variable X has the following probability distribution :A. `(7)/(81)`B. `(5)/(81)`C. `(2)/(81)`D. `(1)/(81)` |
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Answer» Correct Answer - D `therefore sum_(r=0)^(8)P(X=r)=1` `rArr a+3a+5a+7a+9a+11a+13a+15a+17a=1` `rArr a=(1)/(81)` |
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| 15. |
If the probability of a random variable X is a given below : `{:(X=x:,-2,-1,0,1,2,3),(P(X=x):,(1)/(2),k,(1)/(5),2k,(3)/(10),k):}` Then the value of k, is :A. `(1)/(10)`B. `(2)/(10)`C. `(3)/(10)`D. `(7)/(10)` |
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Answer» Correct Answer - A The given distribution is a probability distribution. `therefore P(X=-2)+P(X=-1)+P(X=0)+P(X=1)+P(X=2)+P(X=3)=1` `rArr (1)/(10)+k+(1)/(5)+2k+(3)/(10)+k=1rArr 4k=(4)/(10)rArr k=(1)/(10)` |
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| 16. |
A random, variable `X`has the following probability distribution:`X : 0 1 2 3 4 5 6 7``P(X):0 k 2k 2k 3k k^2 2k^2 7k^2+k`Find each of the following:`k`ii. `P(X |
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Answer» Correct Answer - A Since the sum of the probabilities in a probability distribution is always unity. `therefore P(X=0)+P(X=1)+.....+P(X=7)=1` `rArr 0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1` `rArr 10k^2+9k-1=0` `rArr (10k-1)(k+1)=0` `rArr 10k-1=0 " "[because kge0thereforek+1 ne 0]` `rArr k=(1)/(10)` Now, `P(Xge 6)=P(X=6)+P(X=7)=2k^2+7k^2+k=9k^2+k=(19)/(100)" "(k=1//10]` |
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| 17. |
A dice is thrown `(2n+1)` times. The probability that faces with even numbers show odd number of times isA. `(1)/(2)`B. `lt(1)/(2)`C. `gt(1)/(2)`D. none of these |
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Answer» Correct Answer - A We,have p= Probability of getting an even number in a throw `rArrp=(3)/(6)=(1)/(2)` `therefore q=1-p=(1)/(2)` Let X denote the number of times an even number is shown in `(2n+1)` throws of a dice. Then, X follows binomial distribution such that `P(X=r)=.^2n+1C_(r)((1)/(2))^2n+1-r((1)/(2))^r=.^2n+1C_(r) ((1)/(2))^2n+1`. `therefore` Required Probability `=sum_(r=1)^(2n+1)P(X=r)=sum _(r=1)^(2n+1)2n+1.^C_r((1)/(2))^2n+1` `=((1)/(2))^2n+1{.^(2n+1)C_1+.^(2n+1)C_3+.....+.^(2n+1)C_(2n+1)}` `=(1)/(2^(2n+1))xx2^2n=(1)/(2)`. |
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| 18. |
Two players toss 4 coins each. The probability that they both obtainthe same number of heads is`5//256`b. `1//16`c. `35//128`d. none of theseA. `5//256`B. `1//16`C. `35//128`D. none of these |
| Answer» Correct Answer - C | |
| 19. |
Two dice are tossed 6 times. Then the probability that 7 will show an exactly four of the tosses , isA. `(225)/(18442)`B. `(116)/(20003)`C. `(125)/(15552)`D. none of these |
| Answer» Correct Answer - C | |
| 20. |
A box contains 24 identical balls of which 12 are white and 12 areblack. The balls are drawn at random from the box one at a time withreplacement. The probability that a white ball is drawn for the 4thtime on the 7th draw is`5//64`b. `27//32`c. `5//32`d. `1//2`A. `5//64`B. `27//32`C. `5//32`D. `1//2` |
| Answer» Correct Answer - A | |
| 21. |
A box contains `15` green and `10` yellow balls. If `10` balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is : (a) `12/5` (b) `6` (c) `4` (d) `6/25`A. `(12)/(5)`B. 6C. 4D. `(6)/(25)` |
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Answer» Correct Answer - A Let X denote the number of green balls in a draw 10 balls one-day one with replacement. Then,X is a binomial variate with `n=10,p=(15)/(25)=(3)/(5)and q=(2)/(5)` `therefore` Variance `X=npq=10xx(3)/(5)xx(2)/(5)=(12)/(5)`. |
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| 22. |
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards, then the meanof the number of aces isA. `1//13`B. `3//13`C. `2//13`D. none of these |
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Answer» Correct Answer - C Let X denote the number of aces obtained in two draws, Then,X follows binomial distribution with `n=2`, `p=(4)/(52)=(1)/(13)and q=(12)/(13)` `therefore` Mean of number of aces `=np=(2)/(13)`. |
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| 23. |
There are 12 white and 12 red ball in a bag. Balls are drawn one by one with replacement from the bag. The probability that 7th drawn ball is 4th white, isA. `1//4`B. `5//32`C. `3//16`D. `5//16` |
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Answer» Correct Answer - B We have, `P=` Probability of getting a white ball in a draw `=1//2` `q=` Probability of getting a red ball in a draw `=1//2` Getting `4th` white ball in `7th` draw means that, we must get 3 white balls in the first six draws and a white ball in `7th` draw. `therefore` Required Probability `={.^6C_(3)((1)/(2))^3((1)/(2))^(6-3)}xx(1)/(2)=(5)/(32)` |
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| 24. |
A die is tossed thrice. If event of getting an even number is a success. Then the probability of getting at least two successes isA. `7//8`B. `1//4`C. `2//3`D. `1//2` |
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Answer» Correct Answer - D Let X denote the number of successes in 3 trials. Then, X is a binomial variate with `n=3,p=(3)/(6)=(1)/(2)` such that `P(X=r)=.^3C_(r)((1)/(2))^3,r=0,1,2,3` `therefore` Required Probability `=P(Xge 2)=P(X=2)+P(X=3)` `=.^3C_2((1)/(2))^3+.^3C_(3)((1)/(2))^3=(1)/(2)` |
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| 25. |
Seven chits are numbered 1 to 7. Four chits are drawn one by one with replacment. The probability that the least number appearing on any selected chit is 5 is :A. `((3)/(7))^4`B. `((6)/(7))^3`C. `(5xx4xx3)/(7^3)`D. `((3)/(4))^4` |
| Answer» Correct Answer - A | |
| 26. |
The probability that a candidate secure a seat in Engineering through EAMCET is `1/10`Seven candidate are selected at random from a centre.The probability that exactly two will get seats isA. `15(0.1)^2(0.9)^5`B. `20(0.1)^2(0.9)^5`C. `21(0.1)^2(0.9)^5`D. `23(0.1)^2(0.9)^5` |
| Answer» Correct Answer - C | |
| 27. |
Anexperiment succeeds twice as often as it fails. Find the probability that inthe next six trails there will be at least 4 successes.A. `(64)/(729)`B. `(192)/(729)`C. `(240)/(79)`D. `(496)/(729)` |
| Answer» Correct Answer - D | |
| 28. |
A random variable X takes values `-1,0,1,2` with probabilities `(1+3p)/4,(1-p)/4,(1+2p)/4,(-14p)/4` respectively, where `p` varies over `R.` Then the minimum and maximum values of the mean of `X` are respectivelyA. `-(7)/(4)and (1)/(2)`B. `-(1)/(16)and (5)/(16)`C. `-(7)/(4) and (5)/(16)`D. `-(1)/(16)and (5)/(4)` |
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Answer» Correct Answer - D Since `(1+3p)/(4),(1-p)/(4),(1+2p)/(4)and(1-4p)/(4)` are probabilites when X takes values `-1,0,1` and 2 respectively. Therefore, each is greater tahn or equal to 0 and less than or equal to 1. i.e., `0 le (1+3p)/(4) le 1,0le (1-p)/(4)le 1,0 le (1+2p)/(4)le 1 and 0 le (1-4p)/(4)le 1 rArr -(1)/(3)le p le (1)/(4)` Let `overlineX` be the mean of X. Then, `overline X=-1xx(1+3p)/(4)+0xx(1p)/(4)+1xx(1+2p)/(4)+2xx(1-4p)/(4)` `rArr overline(X) =(2-9p)/(4)` Now, `-(1)/(3)le p le (1)/(4)` `rArr 3 le -9p le -(9)/(4)` `rArr-(1)/(4) le 2-9p le 5rArr -(1)/(16)le (2-9p)/(4)le 5rArr-(1)/(16) le X le (5)/(4)`. |
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| 29. |
If X follows a binomial distribution with parameters `n=100 and p = 1/3`, then `P(X = r)` is maximum whenA. 32B. 34C. 33D. 31 |
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Answer» Correct Answer - C We, have, `(n+1)p=(101)/(3)` which is not an integer. `therefore P(X=r)` is maximum when `r=[(101)/(3)]=33` |
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| 30. |
If X follows a binomial distribution with parameters `n=8` and `p=1//2`, then `p(|X-4|le2)` equalsA. `(118)/(128)`B. `(119)/(128)`C. `(117)/(128)`D. none of these |
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Answer» Correct Answer - B We have, `P(|X-4|le2)=P(-2leX-4le 2)=P(2le X le 6)` `=P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)` `=.^8C_2((1)/(2))^8=+.^8C_3((1)/(2))^8+.^8C_(4)((1)/(2))^8+.^8C_5((1)/(2))^8+.^8C_6((1)/(2))^8` `(1)/(2^8){28+56+70+56+28}=(238)/(2^8)=(119)/(128)` |
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| 31. |
If X is a binomial variate with parameters n and p, where `0gt pgt` such that `(P(X=r))/(P(X=n-r))` is independent of n and r, then p equals.A. `1//2`B. `1//3`C. `1//4`D. none of these |
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Answer» Correct Answer - A We have , `(P(X=r))/(P(X=n-r))=(.^Nc_(r)p^r(1-p)^(n-r))/(.^nC_(n-r)p^(n-r)(1-p)^r)` `rArr (P(X=r))/(P(X=n-r))==((1-p)/(p))^(n-2r)=(1-(1)/(p))^(n-2r)` For this to be independent of n and r, we must have `1-1//p=1rArr p=1//2`. |
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| 32. |
If in a binomial distribution `n=4, P(X=0)=(16)/(81), t h e n P(X=4)`equals`1/(16)`b. `1/(81)`c. `1/(27)`d. `1/8`A. `(1)/(16)`B. `(1)/(81)`C. `(1)/(27)`D. `(1)/(8)` |
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Answer» Correct Answer - B We have, `n=4 and P(X=0)=(16)/(81)` Let p be the probability of success and q that of failure in a trial. Then, `P(X=0)=(16)/(81)` `rArr .^C_(0)q^4=(16)/(81)` ` rArr q^4=((2)/(3))^4rArr q=(2)/(3)rArr p=(1)/(3)` `therefore P(X=4)=.^4C_(4)p^4q^0=p^4=((1)/(3))^4=(1)/(81)` |
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| 33. |
A rifleman is firing at a distance target andhence has only 10% chance of hitting it. Find the number of rounds; he mustfire in order to have more than 50% chance of hitting it at least once.A. 11B. 9C. 7D. 5 |
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Answer» Correct Answer - C Let p be the probability that the rifleman hits the target. Then,`p=(10)/(100)=(1)/(10)and q =(9)/(10)` Suppose n rounds are fired. Let X be the number of times the rifleman hits the target in n trials. Then, `P(X=r)=.^nC_r((1)/(10))^r((9)/(10))^n-r,r=0,1,2,....n` Now, `P(X=ge1)ge (1)/(2)` `rArr 1-P(X=0)ge(1)/(2)` `rArr P(X=0)lt(1)/(2)rArr ((9)/(10))^n le (1)/(2)rArr n=7,8,9.......` |
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| 34. |
A fair coin is tossed 100 times. The probability of getting tails anodd number of times is`1//2`b. `1//8`c. `3//8`d. none of theseA. `1//2`B. `1//8`C. `3//8`D. none of these |
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Answer» Correct Answer - A Let X denote the number of tails. Then X is a binomial variate with parameters `n=100 and p =1//2` such that `P(X=r)=.^100C_(r)((1)/(2))^100,r=0,1,2,...., 100`. `therefore` Required Probability `=P(X=1)+P(X=3)+......+P(X=99)` `=((1)/(2))^100{.^100C_(1)+.^100C_(3)+....+.^100C_(99)}=(1)/(2^100)xx2^99=(1)/(2)`. |
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| 35. |
If two coins are tossed five times, thenthe probability of getting 5 heads and 5 tails isA. `63//256`B. `1//1024`C. `2//205`D. `9//64` |
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Answer» Correct Answer - A Required Probability `=.^10C_(5)((1)/(2))^5((1)/(2))^5=(63)/(256)` |
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