If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is ?

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: 4/13

Explanation:

Number of spades in a standard deck of cards=13
Number of aces in a standard deck of cards=4
And,one of the aces is a spade.
So, 13 + 4 - 1 = 16 spades or aces to choose from.
Therfore,probabiltiy of getting a spade or an ace=16/52=4/13

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: 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

Correct Answer: 1/9

Explanation:

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

=> n(S) = 6 x 6 = 36

E = {(6, 3), (5, 4), (4, 5), (3, 6) }

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9

Correct Answer: 5/9

Explanation:

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in  C 1 6, ways.

 

Now select two distinct number out of remaining 5 numbers which can be done in  C 2 5 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

 

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

  C 1 6 × C 2 5 × 4 ! 2 ! = 720

 =>  n(E) = 720

  ? P E = 720 6 4 = 5 9