Question

A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

Answer 1

When a pair of dice are thrown, then total no. of possible outcomes = 6 × 6 = 36, which are

{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }

E ⟶ event of throwing a no. higher than 9.

No. of favourable outcomes = 6 {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}

We know that P(E) =${\displaystyle \frac{\text{No.of favorable outcomes}}{\text{Total no.of possible outcomes}}}$

i.e., P(E) =6/36 =1/6