199
Q:

# A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

 A) 1/4 B) 1/2 C) 3/4 D) 7/12

Explanation:

Let A, B, C be the respective events of solving the problem and  be the respective events of not solving the problem. Then A, B, C are independent event

are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$∴$ P( none  solves the problem) = P(not A) and (not B) and (not C)

= $PA∩B∩C$

= $PAPBPC$

=  $12×23×34$

= $14$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1-14$= 3/4

Q:

The probability that z lies between 0 and 3.01

 A) 0.4987 B) 0.5 C) 0.9987 D) 0.1217

Explanation:

Probability between z = 0 and z = 3.01 is given by

P(0<z<3.01) = P(z<3.01) - P(z<0)

Reading from the z-table, we have

P(z<0) = 0.5

P(z<3.01) = 0.9987

Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.

0 13
Q:

There are 26 balls marked with alphabetical order A to Z. What is the probability of selecting vowels listed balls?

 A) 1 B) 21/26 C) 5/26 D) 5

Explanation:

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

5 228
Q:

A group of five persons is formed from five boys and four girls. Find the probability that there are at least two girls in the group?

 A) 110/129 B) 5/6 C) 121/126 D) 3/7

Explanation:

Total number of possible ways =

Number of favorable cases =

Therefore, required probability = 105/126 = 5/6

2 331
Q:

Out of 3 girls and 6 boys a group of three members is to be formed in such a way that at least one member is a girl. In how many different ways can it be done?

 A) 64 B) 84 C) 56 D) 20

Explanation:

Total number of possible ways = 9C3 = 84 ways

Required atleast one girl in the group of three = total possible ways - ways in which none is girl

None of the members in the group is girl = 6C3 = 20

Therefore, number of ways that at least one member is a girl in the group of three

= 84 - 20

= 64 ways.

4 319
Q:

A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random, what is the probability that both are red or both are green in colour?

 A) 7 B) 1/7 C) 3/7 D) 0

Explanation:

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability =

1 378
Q:

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?

 A) 35/36 B) 17/36 C) 15/36 D) 1/36

Explanation:

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.

6 877
Q:

In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?

 A) 4/7 B) 2/3 C) 1/2 D) 5/6

Explanation:

Total coins 30

In that,

1 rupee coins 20

50 paise coins 10

Probability of total 1 rupee coins =  20C11

Probability that 11 coins are picked = 30C11

Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.

10 1592
Q:

In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?

 A) 15 B) 18 C) 20 D) 24

Explanation:

We know that, Total probability = 1

Given probability of black stones = 1/4

=> Probability of blue and white stones = 1 - 1/4 = 3/4

But, given blue + white stones =  9 + 6 = 15

Hence,

3/4 ----- 15

1   -----  ?

=> 15 x 4/3 = 20.

Hence, total number of stones in the box = 20.