Question

In a simultaneous throw of a pair of dice, find the probability of getting:

a) 7 as a sum

b) a doublet of odd number

c) not a doublet

d) an odd number on the first dice

Plz give answer in steps

Answer 1

Given :   a simultaneous throw of a pair of dice

To find :  the probability of getting

Solution:

a pair of dice thrown

total outcomes = 36

7 as sum  

( 1 , 6) , ( 2, 5) , ( 3 ,4 ) , ( 4 , 3) , ( 5 , 2) , ( 6 , 1)

Probability of  7 as sum   = 6/36 = 1/6

a doublet of odd number

( 1, 1) , ( 3, 3) , ( 5 , 5)

probability of doublet of odd number  = 3/36 = 1/12

not a doublet

doublet = ( 1, 1) , ( 2, 2) , ( 3, 3) , ( 4, 4) , ( 5, 5) , ( 6, 6)

Probability of  not a doublet = 1  - 6/36  = 5/6

an odd number on the first dice

Probability  = 3/6  = 1/2

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Answer 2

When a simultaneous throw of a pair of dice.

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) and (6, 6).

Total number of possible outcomes = 36

We have to find the probability of getting:

a) 7 as a sum

b) a doublet of odd number

c) not a doublet

d) an odd number on the first dice

Solution:

a) 7 as sum  

All favourable outcomes are:

( 1 , 6), ( 2, 5), ( 3 ,4 ), ( 4 , 3), ( 5 , 2) and ( 6 , 1).

Number of favorable outcomes = 6

∴ P(getting a number 7 as sum) = $\dfrac{6}{36} =\dfrac{1}{6}$

b) a doublet of odd number

All favourable outcomes are:

( 1, 1), ( 3, 3) and ( 5 , 5)

Number of favorable outcomes = 3

∴ P(getting a doublet of odd number) = $\dfrac{3}{36} =\dfrac{1}{12}$

c) not a doublet

Doublet = ( 1, 1), ( 2, 2), ( 3, 3), ( 4, 4), ( 5, 5) and ( 6, 6)

∴ P(getting not a doublet) = 1  -  $\dfrac{6}{36}$ = $\dfrac{5}{36}$

d) an odd number on the first dice

All favourable outcomes are:

1, 3 and 5

Number of favorable outcomes = 3

∴ P(getting an odd number on the first dice) = $\dfrac{3}{36} =\dfrac{1}{12}$