Two dice are rolled simultaneously. Find the probability of getting a total of 9.

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: 1/6

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: 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: 1/2

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: 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: 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

Correct Answer: 21/38

Explanation:

n(S) =  C 2 20 = 190 

n(E) =  C 2 15 = 105 

Therefore, P(E) = 105/190 = 21/38

Correct Answer: 0.043

Explanation:

 Vowels are A I A I O, 

 

 C    A   S    T   I    G   A    T   I    O   N

 

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

 

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

 

n ( E ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600

 

n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600

 

21600 4989600 = 0 . 043

Correct Answer: 1/2

Explanation:

Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2

Correct Answer: 1/2

Explanation:

P(odd) = P (even) = 1 2 1(because there are 50 odd and 50 even numbers)

 

Sum or the three numbers can be odd only under the following 4 scenarios:

 

Odd + Odd + Odd =  1 2 * 1 2 * 1 2=  1 8

 

Odd + Even + Even =  1 2 * 1 2 * 1 2= 1 8

 

Even + Odd + Even =  1 2 * 1 2 * 1 2= 1 8

 

Even + Even + Odd =  1 2 * 1 2 * 1 2 =  1 8

 

Other combinations of odd and even will give even numbers. 

 

Adding up the 4 scenarios above:

 

=  1 8+  1 8+ 1 8+  1 8 =  4 8 =  1 2