Question

a die is thrown twice what is the probability that 1) 15 will not come up either a time? 2) 5 will come up at least once​

Answer 1

1)Total no. Of possible outcomes=36

Favorable outcomes =36

Probability that 15 will not come either time=total favorable outcomes /total possible outcomes

=36/36=1

2)favorable out comes=11

Total possible outcomes =36

Probability that 5 will come up at least once=11/36

hope it will help u

Answer 2

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Outcomes 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)

So, the total number of outcome = 6 × 6 = 36

Method 1:

Consider the following events.

A = 5 comes in the first throw,

B = 5 comes in second throw

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

So, P(notA) = 1 – 6/36 = 5/6

∴ The required probability = 5/6 × 5/6 = 25/36

Method 2:

Let E be the event in which 5 does not come up either time.

So, the favourable outcomes are [36 – (5 + 6)] = 25

∴ P(E) = 25/36

• Number of events when 5 comes at least once = 11 (5 + 6)

∴ The required probability = 11/36

Hope it's Helpful.....:)