Question

A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
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Answer 1

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■Outcomes are:■

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(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

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■(i) Method 1:■

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Consider the following events.

A = 5 comes in 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

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■Method 2:■

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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

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

∴ The required probability = 11/36

Answer 2

Answer:

A die is thrown twice.

So the total number of possible outcomes = 36

(i) Let A be the event that 5 will not come up either time

Total number of possible outcomes for the event A is =36 - 11 = 25

So the required probability = 25/36

(ii) Let B be the event that 5 will come up at least once

Total number of possible outcomes for the event B is = 11

So the required probability = 11/36

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