Probability
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Q.

If five dice are thrown simultaneously, what is the probability of getting the sum as seven?

 A.

\(\dfrac{15}{6^5}\)

 B.

\(\dfrac{11}{6^5}\)

 C.

\(\dfrac{10}{6^5}\)

 D.

\(\dfrac{5}{6^5}\)

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Solution:
Option(A) is correct

The following are two cases when the sum will be 7:

$7=1+1+1+1+3 \Rightarrow ^5C_1$ ways = 5

$7=1+1+1+2+2 \Rightarrow ^5C_2$ ways = 10

Total number of possible ways of throwing five dice \(=6^5\)

The required probability

\(=\dfrac{15}{6^5}\)

Edit: Based on Poonam Pipaliya's comment, solution has been modified.


(5) Comment(s)


Sudhanva Dixit
 ()

How is $1+1+1+2+2=^5C_2$ ways?


Aarti
 ()

It is not mathematically equal, it equals the number of ways three, 1's and two, 2's can be obtained from five dice.

PRAKASH SAMANTA
 ()

${^5C_2}$ ways because-

1+1+1+2+2

1+1+2+1+2

1+2+1+1+2

1+1+2+2+1

1+2+2+1+1

1+2+1+2+1

2+1+1+1+2

2+1+1+2+1

2+1+2+1+1

2+2+1+1+1


Poonam Pipaliya
 ()

in second line there should be $1+1+1+2+2=7$


Deepak
 ()

Yes Poonam, you are right, corrected the solution.