Question

if a pair of dice is thrown once then what is the probability of getting a sum of 8​

Answer 1

Answer:

Number of possible outcomes N(s) = 36.

The favourable outcomes are (sum getting 8) = (2,6) (3,5) (4,4) (5,3) (6,2) (total 5)

Let A be the event of getting a sum of 8.

N(A)=5

P(A)= N(A)/N(S)

= 5/36

Answer is 5/36.

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Answer 2

The probability of getting a sum of 8 is 5/36

Given :

A pair of dice is thrown once

To find :

The probability of getting a sum of 8

Solution :

Step 1 of 3 :

Find total number of possible outcomes

Here it is given that a pair of dice has thrown once

So the event points 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)}

So the total number of possible outcomes = 36

Step 2 of 3 :

Find total number of possible outcomes for the event

Let A be the event that of getting a sum of 8

So the event points are (2,6) (3,5) (4,4) (5,3) (6,2)

So total number of possible outcomes for the event A is 5

Step 3 of 3 :

Find the probability

Hence the required probability

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{5}{36} }$

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