Question

Roll two fair dice. Each die has six faces. Let b be the event that the sum of the two rolls is at most seven. Find p(b).

Answer 1

Answer:

21/36

Step-by-step explanation:

sum 12

(6,6)

1 case

sum 11

(5,6),(6,5)

2 cases

sum 10

3 cases

sum 9

4 cases

sum 8

5 cases

p(b)=1-15/36

21/36

is it brinliest

Answer 2

Therefore the probability of the occurrence of event 'b' p(b) is 7/12.

Given:

2 fair dice are rolled.

'b' is the event that the sum of the 2 rolls is at most 7.

To Find:

The probability of the occurrence of event 'b'.

Solution:

The given question can be solved as shown below.

⇒ The number of possibilities of getting the sum as '2' → ( 1,1 ) → 1 way

⇒ The number of possibilities of getting the sum as '3' → ( 2,1 ); ( 1,2 ) → 2 ways

⇒ The number of possibilities of getting the sum as '4' → ( 2,2 ); ( 1,3 ); ( 3,1 ) → 3 ways

⇒ The number of possibilities of getting the sum as '5' → ( 1,4 ); ( 4,1 ); ( 2,3 ); ( 3,2 ) → 4 ways

⇒ The number of possibilities of getting the sum as '6' → ( 1,5 ); ( 5,1 ); ( 2,4 ); ( 4,2 ); ( 3,3 ) → 5 ways

⇒ The number of possibilities of getting the sum as '7' → ( 1,6 ); ( 6,1 ); ( 2,5 ); ( 5,2 ); ( 3,4 ); ( 4,3 ) → 6 ways

Total number of possibilities of getting the sum '7' at most = 1 + 2 + 3 + 4 + 5 + 6 = 21 ways

Total number of events = 6² = 36 ways

⇒ p(b) = ( Number of possibilities )/(Total number of outcomes )

⇒ p(b) = 21/36 = 7/12

Therefore the probability of the occurrence of event 'b' p(b) is 7/12.

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