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: Throwinga die twice and throwing two dice simultaneously are treated as the same experiment].

Answer 1

Answer:

HEY

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Answer 2

Answer:

(i) 5 will not come up either time

$P(\bar 5)= \frac{25}{36}$

(ii) 5 will come up at least once

$P(5)= \frac{11}{36}$

Step-by-step explanation:

We know that on throwing two dice Total outcomes are 36

(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)

(i) 5 will not come up either time?

Ans. Favourable outcome for 5 will not come up either time:

{(1,1) (1,2) (1,3) (1,4) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,6)

(6,1) (6,2) (6,3) (6,4) (6,6)} =25

Probability of 5 will not come either time

$P(\bar 5)= \frac{25}{36}$

(ii) 5 will come up at least once?

Answer:

Favourable outcome for 5 will come up at least once are:

{(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (1,5) (2,5) (3,5) (4,5) (6,5)}=11

Probability of 5 will come up at least once are: P(5)

$P(5)= \frac{11}{36}$

Or

$P(5)=1-P(\bar 5)\\\\=1 - \frac{25}{36} \\ \\ = \frac{36 - 25}{36} \\ \\ P(5)= \frac{11}{36}$

Hope it helps you