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]​

Answer 1

${\large{\sf{\underline{ Solution:-}}}}$

$\pink{Outcomes\:are:}$

(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

(i) Method 1:

Consider the following events.

A = 5 comes in first throw,

B = 5 comes in second throw

P(A) =$\frac{ 6}{36}$,

P(B) =$\frac{ 6}{36}$ and

P(not B) = $\frac{5}{6}$

So, P(not A) = 1-$\frac{ 6}{36}$ =$\frac{5}{6}$

∴ The required probability = $\frac{5}{6}$×$\frac{5}{6}$ = $\frac{25}{36}$

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) = $\frac{25}{36}$

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

∴ The required probability = $\frac{11}{36}$

Answer 2

Answer:

(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

(i) Method 1:

Consider the following events.

A = 5 comes in first throw,

B = 5 comes in second throw

P(A) =\frac{ 6}{36}

36

6

,

P(B) =\frac{ 6}{36}

36

6

and

P(not B) = \frac{5}{6}

6

5

So, P(not A) = 1-\frac{ 6}{36}

36

6

=\frac{5}{6}

6

5

∴ The required probability = \frac{5}{6}

6

5

×\frac{5}{6}

6

5

= \frac{25}{36}

36

25

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) = \frac{25}{36}

36

25

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

∴ The required probability = \frac{11}{36}

36

11