162
Q:

# Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

 A) 52/221 B) 55/190 C) 55/221 D) 19/221

Explanation:

We have n(s) =$52C2$ 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $52*512*1$ = $26C2$ = 325, n(B)= $26*252*1$= 4*3/2*1= 6  and  n(A∩B) = $4C2$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Q:

A bag contains 2 red Roses, 4 yellow Roses and 6 pink Roses. Two roses are drawn at random. What is the probability that they are not of same color?

 A) 1/6 B) 2/3 C) 14/33 D) 5/6

Explanation:

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

6 337
Q:

What is the probability that a number is divisible by 4?

 A) 2 B) 1/2 C) 1/4 D) 1

Explanation:

Probability is Number of favourable outcomes by total outcomes

Total comes = 2 (Divisible or not)

Favourable = 1 (Divisible)

Hence, probability = 1/2.

2 304
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.

8 596
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.

14 927
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

3 761
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.

7 743
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 =

3 849
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.