Question

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

Answer 1

Answer:

1. 25/36

2. 11/36

Step-by-step explanation:

1. possible outcomes =

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

5,6

6,1

6,2

6,3

6,4

6,5

6,6

total no. of outcomes = 36

total no. of times that 5 will not come up either time = 36-11 = 25

probability = 25/36

2. total no. of outcomes = 36

total no. of times that 5 will  come up at least once = 11

probability = 11/36

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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.....:)