ISI2021-MMA: 2

Question

A person throws a pair of fair dice. If the sum of the numbers on the dice is a perfect square, then the probability that the number $3$ appeared on at least one of the dice is

• A. $1 / 9$
• B. $4 / 7$
• C. $1 / 18$
• D. $7 / 36$

Answer 1

Answer : Option B)  $4/7$

The perfect squares possible are: $4$ and $9$ only with two die.

Total outcomes for getting sum of the numbers on die as a perfect square are:  $\{ (1,3) (2,2) (3,1) (3,6) (4,5) (5,4) (6,3) \}$ = $7$

Favorable outcomes [such that the number 3 appeared on at least one of the die] are : $\{(1,3), (3,1), (6,3), (3, 6)\}$ =  $4$

Hence probability = $4/7$

Answer 2

Answer: (B)

The only possible choices among total 36 choices are (1,3), (3,1), (6,3), (3, 6), (2,2), (4,5), (5,4).
Our favorable outcomes are (1,3), (3,1), (6,3), (3, 6)

Hence probability = $\frac{4}{7}$

Comments

@ankitgupta.1729 Ohh shit right

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@Aditya_  $(3,2), (2,3), (3,3)$ don’t have their sum as perfect squares, how can these be possible choices? The other choices which meet the criteria are $(2, 2), (4, 5), (5, 4)$

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this is conditional  probability  i think.