Test Preparation on Probability For Competitive Exam

1.

A bag contains 8 apple and 6 orange. Four fruits are drawn out one by one and not replaced. What is the probability that they are alternatively of different fruits?

Fruits can be drawn in two format AOAO and OAOA Apple drawn 1st P=8/14*6/13*7/12*5/11 Orange drawn 1st P=6/14*8/13*5/12*7/11 Adding both we get 2[8*7*6*5/14*13*12*11]=2*(10/143)=20/ 143.

2.

In an interview the probability of Praveen to got selected is 0.4. The probability of Geetha to got selected is 0.5. The probability of Sam to got selected is 0.6. The probability of Suresh to got selected is 0.8. What is the probability that at least 2 of them got selected on that day?

Required probability=1 – no one got selected – 1 got selected No one got selected = (1-0.4) x (1-0.5) x (1-0.6) x (1-0.8) = 0.024 1 got selected= 0.4 x ((1-0.5) x (1-0.6) x (10.8) ) + 0.5 x ((1-0.4) x (1-0.6) x (1-0.8)) +0.6 x ((1-0.4) x (1-0.5) x (1-0.8)) + 0.8 x ((1-0.4) x (1-0.5) x (1-0.6)) = 0.016 + 0.024 + 0.036 + 0.096 = 0.17 So, Required probability = 1 – 0.024 – 0.17 = 0.806

3.

A basket contains 10 red ball and 15 white ball. out of which 3 red and 4 white balls are damaged. If two balls selected at random, what is the probability that either both are white balls or both are not damaged?

P(A) = 15c2 / 25c2, P(B) = 18c2 / 25c2 P(A∩B) = 11c2 / 25c2 P(A∪B) = P(A) + P(B) – P(A∩B) => (15c2 / 25c2)+( 18c2 / 25c2)-( 11c2 / 25c2)=406/600==>203/300

4.

A box contains tickets numbered from 1 to 16. 3 tickets are to be chosen to give 3 prizes. What is the probability that at least 2 tickets contain a number which is multiple of 4?

From 1 to 16, there are 4 numbers which are multiple of 4 1st 2 are multiple of 4, and one any other number from (16-4) = 12 tickets 4c2*12c1/16c3 = 72/560 2nd all are multiples of 4. 4c3/16c3=4/560 Add both 72/560+4/560.=76/560

5.

Chance that Sheela tells truth is 35% and for Ramesh is 75%. In what percent they likely to contradict each other in the same question?

P(A) = 35/100=7/20 and P(B) = 75/100=3/4. Now they are contradicting means one lies and other speaks truth. So, Probability = 7/20*1/4 + 13/20 * 3/4 =7/80+39/80=46/80=23/40

6.

Two dice are thrown simultaneously. What is the probability of getting the sum of the numbers as even?

Throw two dice n(s)=36 E is nos sum is even. Hence E={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) , (2,6 ), …………( 6,2 ), (6,4 ), ( 6,6 )} n(E)= 18 Thus required probability= 18/36= 1/2

7.

A basket contains 8 Red and 6 Pink toys. There is another basket which contains 7 Red and 8 Pink toys. One toy is to drawn from either of the two baskets. What is the probability of drawing a Pink toys?

Probability of one basket =1/2 1st Basket Pink toy probability =1/2* (6c1/14c1) 2nd Basket Pink toy probability = 1/2* (8c1/15c1) Adding both (1/2*6/14) + (1/2*4/15) 3/14+4/15=101/210

8.

Four persons are chosen at random from a group of 3 men, 5 women and 4 children. What is the probability of exactly two of them being men?

Total People = 3 + 5 + 4 = 12 n(s)=12c4 Probability of exactly two men and two from others N(e) = 3c2*9c2 ? P= (3c2*9c2)/12c4=>12/55

9.

A box contains 3 ballons of 1 shape, 4 ballons of 1 shape and 5 ballons of 1 shape. Three ballons of them are drawn at random, what is the probability that all the three are of different shape?

Total=3+4+5=12 n(s)=12c3=220 n(e)=3c1 * 4c1 * 5c1 =60 p=60/220=3/11

10.

12 persons are seated around a round table.What is the probability that two particular persons sit together?

In a circle of n different persons, the total number of arrangements possible = (n – 1)! n(S) = (12 – 1) = 11 ! Taking two persons as a unit, total persons = 11 Therefore no. of ways for these 11 persons to around the circular table = (11 – 1)! = 10! In any unit, 2 particular person can sit in 2! ways. Hence total number of ways that any three person can sit, =n(E) = 10! * 2! Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = (10! * 2!)/11! = 2/11.


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