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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. |
If an event cannot occur, then its propbability isA. 1B. `(3)/(4)`C. `(1)/(2)`D. 0 |
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Answer» Correct Answer - D The event which cannot occur is said to be impossible event and probability of impossible event is zero. |
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| 2. |
The probability that bulbs selected randomly from the lot has life less than 900 h, isA. `11/40`B. `5/16`C. `7/16`D. `9/16` |
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Answer» Correct Answer - D Total number of bulbs in a lot, n(S)=80 Number of bulbs whose life time is less than 900h, n(E ) =10 + 12 + 23=45 `therefore` Probability that bulbs has life time less than 900 h`=(n(E ))/(n(S))=45/80=9/16` Hence, the probability that bulb has life time less than 900 is `9/16`. |
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| 3. |
If the probability of an event is P, then the probability of its completmentry event will beA. P-1B. PC. 1-PD. `1-(1)/(P)` |
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Answer» Correct Answer - C Since probability of am event+probability of its complementry event=1 So, probability of its complementry event=-1Probability of an event=1-P |
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| 4. |
An event is very unlikely to happen. Its probability is closet toA. 0.0001B. 0.001C. 0.01D. 0.1 |
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Answer» Correct Answer - A The probability of an event which is ver unlikely to happen is closest to zero and from the given option 0.0001 is closet to zero. |
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| 5. |
Which of the following cannot be the probability of an event?A. `(1)/(3)`B. 0.1C. 3D. `(17)/(16)` |
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Answer» Correct Answer - D Since, probability of an event always lies between 0 and 1 |
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| 6. |
Can the experimental probability of an event be a negative number? If not, why? |
| Answer» No, since the number of trials in which the event can happen cannot be negative and the total number of trials is always positive. | |
| 7. |
The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot isA. 7B. 14C. 21D. 28 |
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Answer» Correct Answer - B Here, total number of eggs=400 Probability of getting a bad egg=0.035 `Rightarrow ("Number of bad eggs")/("Total number of eggs")=0.0035` `Rightarrow =("Number of bad eggs")/("400")=0.035` `therefore "Number of bad eggs"=0.035xx400=14` |
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| 8. |
If P(A) denotes the probability of an event, thenA. `P(A) lt 0`B. `P(A) gt 1`C. `0 le P (A) le 1`D. `-1 le P(A) le 1` |
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Answer» Correct Answer - C Since, probability of an event always lies between 0 and 1 |
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| 9. |
Refer to 0.28. What is the probablity that the card is (i) a club (ii) 10 of hearts |
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Answer» (i) Let `E_(3)`=Event of getting a club `n(E_(3))`=(13-3)=10 `therefore "Required probability"=(n(E_(3)))/(n(S))=(10)/(49)` (ii) Let `E_(4)`=Event of getting 10 g of hearts `n(E_(4)=1` [Because in 52 playing cards only 13 are the heart cards and only one 10 in 13 heart cards] `therefore "Required probability"=(n(E_(4)))/(n(S))=(1)/(49)` |
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| 10. |
If a card is selected from a deck of 52 cards, then the probability of a being a red face card isA. `(3)/(26)`B. `(3)/(13)`C. `(2)/(13)`D. `(1)/(2)` |
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Answer» Correct Answer - A In a deck of 52 cards, there are 12 face cards, i.e. face cards i.e. 6 red and 6 black cards. So, probability of getting a red face card=`(6)/(52)=(3)/(26)` |
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| 11. |
A card is drawn from a dect of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favorable to E isA. 4B. 13C. 48D. 51 |
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Answer» Correct Answer - D In a dect of 52 cards, there is 13 cards of heart and 1 is ace of heart. Hence, the number of outcomes favourable to E=51 |
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| 12. |
Cards with number 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has (i) an even number (ii) a square number |
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Answer» Total number of out comes with number 2 to 101, n(s)=100 (i) Let `E_(1)` =Event of selecting a card which is even number ={(2,4,6…10)} `["in an AP",l=a+(n-2)d, here l=100, a=2 and d=2 Rightarrow 100=2+(n-1)2 Rightarrow (n-1)=49 Rightarrow n=50]` `therefore n(E_(1))=50` `therefore "Required probability"= (n(E_(1))/(n(S))=(50)/(100)=(1)/(2))` (ii) Let `E_(2)`=Event of selecting a card which is a square number {(4,16,25,36,49,64,81,100)} `{(2)^(2),(3)^(2),(4)^(2),(5)^(2),(6)^(2),(7)^(2),(8)^(2),(9)^(2),(10)^(2)}` `therefore n(E_(2))=9` Hence required probability=`(n(E_(2)))/(n(S))=(9)/(100)` |
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| 13. |
A recent survey found that the edges ages of workers in a factory as follow. If a person is selected at random, find the probability that the person isA. 40 yr or moreB. under 40 yrC. having age from 30-39 year.D. under 60 but over 39 year. |
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Answer» Total number of worker in a factory. n(S)=38+27+86+46+3=200 i) Number of persons selected at the edge of 40 yr or more, `P(E_(1))=86+46+3=135` Probability that the persons selected at the age of 40 yr or more, `P(E_(1))=(n(E_(1)))/(n(S))=135/200=0.675` Hence, the probability that the person selected at the age of 40 yr or more is 0.675. ii) Number of persons selected under the age of 40 yr `(n(E_(2)))=38+27=65` `therefore` Probability that the persons selected under the age of 40 yr. `P(E_(2))=n(E_(2))/n(S)=65/200=0.325` Hence, the probability that the person selected under the age of 40 yr is 0.325. iii) Number of persons selected having age from 30 to 39 yr, `n(E_(3))=27` `therefore` Probability that the person selected having age from 30 to 39 yr, `P(E_(3))=(n(E_(3)))/(n(S))=27/200=0.135` Hence, the probability that the person selected having age from 30 to 39 yr is 0.135. iv) Number of persons selected having age under 60 but over 39 yr. `n(E_(4))=86+46=132` `therefore` Probability that the person selected having age under 60 but over 39 yr. `P(E_(4))=(n(E_(4)))/(n(S))=132/200=0.66` Hence, the probability that the person selected having age under 60 but over 39 yr is 0.66 |
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| 14. |
One ticket is drawn at random from a bag containing ticket numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is.A. `(1)/(5)`B. `(3)/(5)`C. `(4)/(5)`D. `(1)/(3)` |
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Answer» Correct Answer - A Number of total outcomes=40 Multiples of 5 between 1 to 40=5,10,15,20,25,30,35,40 `therefore` Total number of possible outcomes=8 `therefore Required=(8)/(40)=(1)/(5)` |
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| 15. |
A company selected 4000 households at random and surveyed them to find out a relationship between income level and the number of televisions sets in a home. The information, so obtained is listed in the following table. Find the probability i) of a household earning Rs 10000- Rs 14999 per year and having exactly one television. ii) of a household earning Rs 25000 and more per year owning 2 televisions. iii) of a household not having any television. |
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Answer» The total number of the households selected by the company. n(S)=4000 i) Number of households earning Rs 10000 - Rs 14999 per year and having exactly one television, `n(E_(1))=240` `therefore` Required probability =`(n(E_(1)))/(n(E_(S)))=240/4000=6/100=3/50=0.06` Hence, the probability of a household earning Rs 1000 - Rs 14999 per year owning 2 televisions, `n(E_(2))=760` Required probability = `(n(E_(2)))/(n(S))=760/4000` `=0.19` Hence, the probability of a household earning Rs 25000 and more per year owning 2 televisions is 0.19. Hence, the probability of a household earning Rs 25000 and more per year owning 2 televisions is 0.19. iii) Number of households not having any televisions, `n(E_(3))=30` `therefore` Required probability`=(n(E_(3)))/(n(S))= 30/4000=3/400` Hence, the probability of a household not having any television is `3//400`. |
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| 16. |
Two dice are thrown simultaneously. What is the probability that the sum of the number appearing on the dice is (i) 7? (ii) a prime number? (iii) 1? |
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Answer» Two dice are thrown simultaneously So, total number of possible outcomes=36 (i) Sum of the number appearing on the dice is 7. So, the possible ways are (1,6),(2,5),(3,4),(4,3),(5,2) and (6,1) Number of possible ways=6 `therefore` Required probability=`(6)/(36)=(1)/(6)` (ii) Sum of the numbers appearing on the dice is a prime number i.e. 2,3,5,7 and 11 So, the possible ways are (1,1), (1,2), (2,1), (1,4), (2,3),(3,2), (4,1), (1,6), (2,5),(3,4),(4,3),(5,2),(6,1),(5,6) and (6,5) Number of possible ways=15 `therefore` "Required probability=(15)/(36)=(5)/(12)` (iii) Sum of the number appearing on the dice is 1. It is not possible, so its probability is zero. |
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| 17. |
Two dice are thrown simultaneouly 500 times. Each time the sum of the two numbers appearing on their tops is noted and recorded as given in the following table If the dice are thrown once more, then what is probability of getting a sumA. 3?B. more than 10?C. less than or equal to 5?D. between 8 and 12? |
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Answer» Total number of times, when two dice are thrown simulatenously, n(S)=50 i) Number of times of getting a sum 3, n( E)=30 `therefore` Probabilty of getting a sum 3 =`(n( E))/(n(S))=30/500=3/50=0.06` Hence, the probability of getting a sum 3 is 0.06. ii) Number of times of getting a sum more than 10, `n(E_(1))=28 + 15=43` `therefore` Probability of getting sum more tan 10 =`(n(E_(1)))/(n(S))= 43/500=0.086` Hence, the probability of getting a summore than 10 is 0.086 iii) Number of times of getting a sum less than or equal to 5, `n(E_(2))=55+42+30+14=14` `therefore` Probability of getting a sum less than or equal to 5 is 0.282. iv) The number of times of getting a sum between 8 and 12, `n(E_(3))=53+46+28=127` `therefore` Required probability`=(n(E_(3)))/(n(S))=127/500=0.254` Hence, the probability of getting a sum between 8 and 12 is 0.254. |
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| 18. |
Two dice are thrown at the same time. Determine the probability that the difference of the number on the two dice is 2. |
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Answer» Total number of sample space in two dice, n(S)=`6xx6=36` Let, E=Event of sample the number whose difference is 2 ={(1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4)} `therefore n(E )=8` `therefore P(E )=(n(E ))/(n(S))=(8)/(36)=(2)/(9)` |
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| 19. |
The weights of coffee im 70 packets are shown in the following table Determine the modal weight. |
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Answer» In the given data, the highest frequency is 26, which is lies in the interval 201-202 Here, `l=201, f_(m)=26, f_(1)=12, f_(2)=20 and ("class width")h=1` `therefore "Mode"l+((f_(m)-t_(1))/(2f_(m)-t_(1)-f_(2)))xxh=201+((26-12)/(2xx26-12-20))xx1` `=201+((14)/(52-32))=201+(14)/(20)=201+0.7=201.7g` Hence the modal weight is 201.7g |
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| 20. |
A school has five houses A, B,C,D and E. A class has 23 students, 4 from houses A, 8 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C isA. `(4)/(23)`B. `(6)/(23)`C. `(8)/(23)`D. `(17)/(23)` |
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Answer» Correct Answer - B Total number of students=23 Number of students in house A,B and C =4+8+5=17 `therefore` Remains students=23-17=6 So, probability that the selected is not from A, B and C=`(6)/(23)` |
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| 21. |
The mean of 25 observation is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation isA. 23B. 36C. 38D. 40 |
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Answer» Correct Answer - B Given, mean of 25 observations = 36 `therefore` Sum of 25 observations `=36 xx 25=900` Now, the mean of first 13 observations`=13 xx 32=416` and the mean of last 13 observations =40 `therefore` sum of last 13 observations = (Sum of last 13 observations + Sum of first 13 observations)-(Sum of 25 observations) `=(520-416)-900=936-900=36` Hence, the 13th observation is 36. |
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| 22. |
While computing mean of grouped data, we assume that the frequecies areA. evenly distributed over all the classesB. centred at the class marks of the classesC. centred at the upper limits of the classesD. centred at the lower limits of the classes |
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Answer» Correct Answer - B In computing the mean of grouped data, the frequencies are centred at the class marks of classes. |
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| 23. |
There are 100 sealed envelops in a box, 10 of them contain a cash prize of 100 each, 100 of them contain a cash prize of 50 each and 200 of them contain a cash prize of 10 each and rst do not contin any cash prize. If they are well shuffled and an evelope is picket up out, What is the probabiilty that it contains no cash prize? |
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Answer» Total number of sealed envelopes in a box, n(S)=1000 Number of evelopes containing cash prize=10+100+200=310 Number of envelops containing no cash prize, `n(E )=1000-310=690` `P(E )=(n(E ))/(n(S))=(690)/(1000)=(69)/(100)=0.69` |
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| 24. |
A game consists of spining an arrow which comes to rest pointing at one of regions (1,2 or 3) (see figure). Are the outcomes 1,2 and 3 equally likely to occur? Give reason |
| Answer» No, the outcomes are not equally likely, because 3 contains half part of the total region, so it more likely than 1 and 2, since 1 and 2, each contains half part of remaining part of the region. | |
| 25. |
The mean marks (out of 100) of boys and girls in an examinatin are 70 and 73, respectively. If the sum mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls. |
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Answer» Let x and y be the number of boys and girls, respectively. Given, mean marks (out of 100) of boys `(barx_(1))=70` and mean marks (out of 100) of girls `(barx_(2))` =73 Also, given that, mean marks of all the students in the exmaination `(barx_(12))=71` Now, using the formula, Combined mean , `(barx_(12))= (barx_(1)n_(1)+barx_(2)n_(2))/(n_(1)+n_(2))=71` (Given) `therefore (70n_(1) + 73n_(2))/(n_(1)+n_(2))=71` `rArr 70n_(1)+73n_(2)= 71n_(1)+71n_(2)` `rArr 73n_(2)-71n_(2)=71n_(1)-70n_(1)` `rArr 2n_(2)=n_(1)` `rArr n_(1)/n_(2)=2/1 or n_(1):n_(2)=2:1` |
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| 26. |
The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class isA. 15B. 25C. 35D. 40 |
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Answer» Correct Answer - C Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class =5 `rArr x-y=5` ……….(i) Also, given lower class (y)=10 On putting y=10 in Eq. (i), we get `x-10=5 rArr x=15` So, the upper class limit of the lowest class is 15. Hence, the upper limit of the highest class = (Number of continuous classes x Class with + Lower class limit of the lowest class) `=5 xx 5 + 10 = 25+10=35` Hence, the upper class limit of the highest class is 35. Alternate method After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15, 15-20, 20-25, 25-30, and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of the class is 35. |
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| 27. |
The mole of given data 15,14,19.20,14,15,16,14,15,18,14,19, 15,17 and 15 isA. 14B. 15C. 16D. 17 |
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Answer» Correct Answer - B We first arrange the given data in ascending ordrer as follows 14,14,14,14,15,15,15,15,15,16,17,18,19,19,20` From above, we see that 15 occurs most frequently i.e., 5 times. Hence, the mode of the given data is 15. |
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| 28. |
The class marks of a frequency distribution are given as follows 15,20, 25,……….. The class corresponding to the class mark 20 isA. `12.5-17.5`B. `17.5-22.5`C. `18.5-21.5`D. `19.5-20.5` |
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Answer» Correct Answer - B Since, the difference between mid values is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option ( c) and (d) are wrong. Since, the mid values is 20 which can get only, if we take option (b) `(17.5 + 22.5)/2=40/2=20` |
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| 29. |
The range of the data 25,18,20,22,16,6,17,15,12,30,32,10,19,8,11 and 20 isA. 10B. 15C. 18D. 26 |
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Answer» Correct Answer - D In a given data, maximum value = 32 and minimum value =6 We know, range of the data= maximum value - minimum value `=32-6=26` Hence, the range of the given data is 26. |
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| 30. |
The mark of the class `90-120` isA. 90B. 105C. 115D. 120 |
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Answer» Correct Answer - B In a given class `90-120`, upper class =120 and lower class=90 We know that, class mark `=("Upper class + Lower class")/(2)` `=(120+90)/(2)=210/2=105` |
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| 31. |
The times(in second) taken by 150 atheletes to run a 110 m hurdle race are tabulated below The number of atheletes who completed the race in less than 14.6s isA. 11B. 71C. 82D. 130 |
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Answer» Correct Answer - C The number of atheletes who completed the race in less than 14.6=2+4+5+71=82 |
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| 32. |
As the number of tosses of a coin increases the ratio of the number of heads to the total number of tosses will be`1/2`. Is it correct? If not, write the correct one. |
| Answer» No, since the number of coin increases, the ratio of the number of heads to the total number of tosses will be nearer to `1/2` but not exactly `1/2`. | |
| 33. |
Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons. |
| Answer» The outcome of next toss or may not be tail, because on tossing a coin, we get head or tail so both are equally likely events. | |
| 34. |
If I toss a coin 3 times and get head and get head each time, should I expect a tail to have a higher chance in the 4th toss? Give a reason in support of your answer. |
| Answer» No, let we toass a coin, then we get head or tail, both are equally events. i.e. probability of each events is `(1)/(2)`. So, no question of expecting a tail to have a higher chance in 4th toss. | |
| 35. |
I toss three coins together. The possible outocmes are no heads, 1 head 2 head and 3 heads. So, I say that prbability of no heads is `(1)/(4)`. What is wrong with this conclusion? |
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Answer» I toss three coins together So, total number of outcome `=2^(3)=8` and possible outcomes are (HHH),(HTT),(THT),(TTH),(HHT),(THH),(HTH),and (TTT) Now, probability of getting no head=`(1)/(8)` Hence, the given conclusion is wrong because the probability of no head is `(1)/(8) not (1)/(4)` |
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| 36. |
If you toss a coin 6 times it comes down head on each occasion Can you say that the probability of getting a head is 1? Give reasons |
| Answer» No, if let we toss a coin, then we get head or tail, both are equally likely events. So probability is `(1)/(2)`. If we toss a coin 6 times, then probability will be same in each case. So, the probability of getting a head is not 1. | |
| 37. |
Two coins are tossed 1000 times and the outcomes are recorded as below Based on this information, the probability for atmost one head isA. `1/5`B. `1/4`C. `4/5`D. `3/4` |
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Answer» Correct Answer - C The total number of coins tossed, n(S)=1000 Number of outcomes in which atmost one head, n( E)=550+250=800 `therefore` Probability for atmost one head =`(n( E))/(n(S))=800/1000=4/5` Hence, the probability for atmost one head is `4/5`. |
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| 38. |
A coin is tossed two times. Find the probaility of getting atmost one head. |
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Answer» The possible outcomes, If a coin is tossed 2 times is S={(HH),(TT),(HT),(TH)} Let E=Event of getting at most one head ={(TT),(HT),(TH)} n(E )=3 Hence, required probability=`(n(E ))/(n(S))=(3)/(4)` |
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| 39. |
In the class intervales 10-20, 20-30, the number is 20 is included inA. 43758B. 20-30C. Both the intervalsD. None of these |
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Answer» Correct Answer - B Since, the class interval 10-20 is the first interval of frequency distribution and 20-30 is the next one but the number 20 is present in both intervals. We know that, the presence of 20 in the interval 10-20 is not fully `100%` while in the next interval `20-30`, presence of it fully `100%`. |
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| 40. |
A child says that the median of 3,14, 18,20 and 5 is 18. What doest not the child understand about finding the median? |
| Answer» The child does not understand, that data has to be arranged in ascending or descending order before finding the median. | |