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

A dice is thrown twice. The possibilites 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)

We have to find the probability of getting 5 will not come up either time and 5 will come up at least once.

We know that,

Probability = (Number of favourable outcomes)/(Total number of outcomes)

i) Total number of outcomes = 6*6 = 36

Number of favourable outcomes = 25

(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)

P(5 will not come up either time) = 25/36

ii) Total number of favourable outcomes = 36

Number of favourable outcomes = 11

(1,5) (2,5) (3,5) (4,5) (6,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

P(5 will come up at least once) = 11/36

Answer 2

ANSWER

CONCEPT USED

throwing a die twice and throwing two dice simultaneously are treated as the same experent

EXPLANATION

(1) 5 will not come up either time

= 5 will not come up in first attemt × five will not come up in second attempt

= P{(not5) (not5) }= p(not5) P(not5)

here u use conditional probability

and 5 not come in second attemt is independent of first attempt

= 5/6×5/6=25/36

(2)5 will come at least once

1 throw 2 throw

5come 5 not come

5 not come 5come

5come 5 come

so P(at least once come)= 1/6×5/6+ 5/6×1/6+1/6×1/6= 11/36