Question

A single die is thrown twice. what is the probability that the sum is neither 8 nor 9?

Answer 1

A single die is thrown twice
Let 'S' be the sample space
S = { (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)}

n(S) = 36
Let Event A be the sum of digits on the upper face is neither 8 nor 9

A = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (3,1),
(3,2), (3,3), (3,4), (4,1), (4,2), (4,3),
(4,6), (5,1), (5,2), (5,5), (5,6), (6,1),
(6,4), (6,5), (6,6)}
n(A)= 27
p(A)= n(A)/n(S) = 27/ 36 = 3/4

Ans : The probability that the sum is neither 8 nor 9 is 3/4

I hope this helps you.

Answer 2

P(getting the sum is neither 8 nor 9)$=\dfrac{3}{4}$

Step-by-step explanation:

When a single die is thrown twice:

All possible 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)}

Total number of possible outcomes = 36

Let E be the event of sum of digits on the upper face is  8 or 9.

The favourable outcomes are:

(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (5,3), (5,4), (6,2) and (6,3).

Number of favourabe outcomes = 9

$P(E)=\dfrac{Number of favorable outcomes}{Total number of possible outcomes}$

$=\dfrac{9}{36}$

P(getting the sum is neither 8 nor 9) = 1 - P(E)

$=1-\dfrac{9}{36} =\dfrac{36-9}{36}$

$=\dfrac{27}{36}=\dfrac{3}{4}$

Hence, P(getting the sum is neither 8 nor 9)$=\dfrac{3}{4}$