#### Question

In a simultaneous throw of a pair of dice, find the probability of getting a doublet

#### Solution

In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) 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 getting a doublet

No. of favorable outcomes = 5 {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}

Total no. of possible outcomes = 36

We know that, Probability P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"`

P(E) =6/36 =1/6

Is there an error in this question or solution?

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In a Simultaneous Throw of a Pair of Dice, Find the Probability of Getting A Doublet Concept: Probability - A Theoretical Approach.

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