Question

Two dice are thrown simultaneously. What is the probability that:
(i)5 will not come up on either of them?
(ii)5 will come up on at least one?
(iii)5 will come up at both dice?

Answer 1

SOLUTION :

Given : Two dice are thrown simultaneously  

If we throw two dices then there possible outcomes are as follows:

{(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)}

Total no. of possible outcomes when 2 dice are thrown = 6 x 6 = 36

(i) Let E1 = Event of 5 will not come up on either of them

Favorable outcomes(5 not coming up on either of them) :  

{(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)}

No. of favorable outcomes = 25  

Probability ,P(E1) = Number of favourable outcomes / total number of outcomes

P(E1) = 25/36

Hence, the Probability that  5 will not come up on either of them, P(E1) = 25/36

(ii) Let E2 = Event of 5 will come up on at least once

Favorable outcomes (5 will come up on at least once) : {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) (6, 5)}

Number of favorable outcomes = 11

Probability ,P(E2) = Number of favourable outcomes / total number of outcomes

P(E2) = 11/36

Hence, the Probability that  5 will come up on at least once, P(E2) = 11/36

(iii)  Let E3 = Event of getting 5 on both dice

Favorable outcomes(5 on both dice) : {(5, 5)}

Number of favorable outcomes = 1

Probability ,P(E3) = Number of favourable outcomes / total number of outcomes

P(E3) = 1/36

Hence, the Probability that  5 will come on both dice, P(E3) = 1/36

HOPE THIS ANSWER WILL HELP  YOU….

Answer 2

Hey mate here's the solution...


Two dice thrown,

So, the total outcomes = 6² = 36

1. 5 will not come either of them

Favourable Outcomes = 36-12 = 24

P( not e) = Favorable outcomes/total Outcomes

= 24/36 = 2/3

2. 5 will come at least once

Favorable outcomes = 12

P(e) = Favorable outcomes/total Outcomes

=12/36 = 1/3

3. 5 will come both twice

Favourable Outcomes =2

P(e) = Favorable outcomes/total Outcomes

= 2/36

= 1/18


Hopefully it helps you...