Probability Questions Quiz 1

Explanation:

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) = n(E)/n(S) = 2/52 = 1/26

Q2. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A. 1/33
B. 3/41
C. 2/91
D. 1/30

Answer: Option C

Explanation:

Let S be the sample space.

Then, n(S)= number of ways of drawing 3 balls out of 15

= ¹⁵C₃

=(15 x 14 x 13)/(3 x 2 x 1)= 455.

Let E = event of getting all the 3 red balls.

n(E) = ⁵C₃ = ⁵C₂ =(5 x 4)/(2 x 1)= 10.

P(E) = n(E)/n(S)= 10/455= 2/91

Q3. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A. 1/13
B. 3/12
C. 40/41
D. 13/102

Answer: Option D

Explanation:

Let S be the sample space.

Then, n(S) = ⁵²C₂ =(52 x 51) / (2 x 1)= 1326.

Let E = event of getting 1 spade and 1 heart.

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13

= (¹³C₁ x ¹³C₁)

= (13 x 13)

= 169.

P(E) =n(E)/n(S)= 169/1326 = 13/102

Q4. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A. 1/13
B. 3/13
C. 2/11
D. 3/11

Answer: Option B

Explanation:

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12 / 52 = 3/13

Q5. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A. 3/8
B. 4/7
C. 2/5
D. 1/6

Answer: Option B

Explanation:

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8/14 = 4/7

Q6. A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2

A. 5/7
B. 3/7
C. 3/5
D. 1/3

Solution:
Option( B) is correct

Explanation :

|X| can take 7 values.
To get |X|<2 ( i.e., −2<X<+2) take X={−1,0,1}

= P(|X|<2)= Favourable CasesTotal Cases
= 3/7

Show Explanation

Hide Explanation

Q7. Two brother X and Y appeared for an exam. Let A be the event that X is selected and B is the event that Y is selected.

The probability of A is 17 and that of B is 29. Find the probability that both of them are selected.

1/63
2/35
2/63
9/14

Solution:
Option( C) is correct

Explanation :

Given, A be the event that X is selected and B is the event that Y is selected.

P(A)=17, P(B)=29.

Let C be the event that both are selected.

P(C)=P(A)×P(B) as A and B are independent events:

=(17)×(29)

= 2/63

Q8. An urn contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, what is the probability that both are red?

A. 6/13
B. 5/26
C. 5/13
D. 7/26

Solution:
Option( B) is correct

Explanation : P(Both are red),

=6C213C2

=5/26

Q9. Four dice are thrown simultaneously. Find the probability that all of them show the same face.

A. 1/216
B. 1/36
C. 4/216
D. 3/216

Solution:
Option( A) is correct

Explanation :

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

=6×6×6×6=64

n(S)=64

Let X be the event that all dice show the same face.

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

n(X)=6

Hence required probability,

=n(X)n(S)=664

=1/216

Q10. A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement

What is the probability that first is white and second is black?

A. 18/145
B. 18/29
C. 36/135
D. 36/145

Solution:
Option( D) is correct

Explanation :

The probability that first ball is white:

=12C130C1

=1230

=25

Since, the ball is not replaced; hence the number of balls left in bag is 29.

Hence, the probability the second ball is black:

=18C 129C1

=1829

Required probability,

=(25)×(1829)

=36145