If a card is drawn at random from a pack of 52 cards,what is the chance of getting a spade or ace?

Correct Answer: 5:4

Explanation:

Here,P(A)=4/9=p/(p+q)

P(odd against event A)=q/p=5/4

Correct Answer: 112/495

Explanation:

n S = C 4 12 = 495

 

n E = C 2 8 × C 1 4 = 112

 

? P ( E ) = 112 495

Correct Answer: 1/2

Explanation:

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
? The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

? required probability = 26/52 = 1/2.

Correct Answer: 7/44

Explanation:

Here, n(E) =  7 C 1 × 5 C 1 × 5 C 1

 

And,  n(S) =  12 C 1 * 11 C 1 * 10 C 1

P(S) =  7 * 6 * 5 12 * 11 * 10 = 7/44

Correct Answer: 2/5

Explanation:

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) +  (Prob. that brother is not selected) × (Prob. that sister is selected)

 

=  1 5 * 2 3 + 4 5 * 1 3=  2 5

Correct Answer: 1/3

Explanation:

In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

Correct Answer: 1/2

Explanation:

The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.

Correct Answer: 5/21

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (?C? x ?C?)

As remaining person can be any one among three couples left.

Required probability = (?C? x ?C?)/¹?C?
= (10 x 6)/252 = 5/21

Correct Answer: 1/6

Explanation:

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36

 

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

  P ( E ) = n ( E ) n ( S ) = 6 36 = 1 6

Correct Answer: 110/129

Explanation:

Total number of possible ways = 

9 C 5 = 9 x 8 x 7 x 6 x 5 5 x 4 x 3 x 2 x 1 = 126

Number of favorable cases = 

$$\begin{gathered} 4 C 2 x 5 C 3 + 4 C 3 x 5 C 2 + 4 C 4 x 5 C 1 \\ 4 x 3 2 x 1 x 5 x 4 x 3 3 x 2 x 1 + 4 x 3 x 2 3 x 2 x 1 x 5 x 4 2 x 1 + 1 x 5 \\ = 6 x 10 + 4 x 10 + 5 \\ = 60 + 40 + 5 \\ = 105 \end{gathered}$$

 

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