In: Statistics and Probability
Find the probability that the sum is as stated when a pair of dice is rolled. (Enter your answers as fractions.) (a) odd and less than 5 (b) odd or less than 5
Solution:
We are given that: two dice are rolled.
Thus sample space of rolling of two dice is:
X = sum of numbers obtained on uppermost faces of two dice
then following is the table which gives all possible values of sum of numbers and their frequencies.
For example: (1,1) gives sum = 1+1=2 ,
(1,2) gives sum = 1+2=3
and so on.
We get:
Die 1 | Die 2 | Sum |
1 | 1 | 2 |
1 | 2 | 3 |
1 | 3 | 4 |
1 | 4 | 5 |
1 | 5 | 6 |
1 | 6 | 7 |
2 | 1 | 3 |
2 | 2 | 4 |
2 | 3 | 5 |
2 | 4 | 6 |
2 | 5 | 7 |
2 | 6 | 8 |
3 | 1 | 4 |
3 | 2 | 5 |
3 | 3 | 6 |
3 | 4 | 7 |
3 | 5 | 8 |
3 | 6 | 9 |
4 | 1 | 5 |
4 | 2 | 6 |
4 | 3 | 7 |
4 | 4 | 8 |
4 | 5 | 9 |
4 | 6 | 10 |
5 | 1 | 6 |
5 | 2 | 7 |
5 | 3 | 8 |
5 | 4 | 9 |
5 | 5 | 10 |
5 | 6 | 11 |
6 | 1 | 7 |
6 | 2 | 8 |
6 | 3 | 9 |
6 | 4 | 10 |
6 | 5 | 11 |
6 | 6 | 12 |
Thus frequency distribution is:
X = Sum of Numbers | Frequency |
2 | 1 |
3 | 2 |
4 | 3 |
5 | 4 |
6 | 5 |
7 | 6 |
8 | 5 |
9 | 4 |
10 | 3 |
11 | 2 |
12 | 1 |
N = 36 |
Part a) P( Sum is odd and less than 5 ) =................?
From above frequency table, we can see: Sum less than 5 are ( 2,3,4) and out of which odd sum is only 3
and sum =3 has frequency 2
Thus we have to find:
P( Sum is odd and less than 5 ) = P( Sum = 3)
P( Sum is odd and less than 5 ) = 2 / 36
P( Sum is odd and less than 5 ) = 1/18
Part b) P(Sum is odd or less than 5 )=........?
P(Sum is odd or less than 5 )=P( Sum is Odd ) + P( Sum is less than 5) - P( Sum if odd and less than 5)
From given frequency distribution, we can see:
Sum is odd has following outcomes: ( 3, 5, 7, 9, 11) with corresponding frequencies ( 2, 4, 6, 4, 2)
Thus
P(Sum is odd) = (2+4+6+4+2) / 36
P(Sum is odd) = 18/36
and
Outcomes of Sum is less than 5 are ( 2, 3, 4) with frequencies ( 1 , 2 , 3 )
Thus
P( Sum is less than 5 ) = (1+2+3)/36
P( Sum is less than 5 ) = 6/36
and
From part a) we have:
P( Sum is odd and less than 5 ) = 2 / 36
Thus
P(Sum is odd or less than 5 )=P( Sum is Odd ) + P( Sum is less than 5) - P( Sum if odd and less than 5)
P(Sum is odd or less than 5 )= 18/36 + 6/36 - 2/36
P(Sum is odd or less than 5 )= (18+6-2) / 36
P(Sum is odd or less than 5 )= 22/36
P(Sum is odd or less than 5 )= 11 / 18