A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?

Question:

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?

(i) that the sum of the two numbers that turn up is 7?

(ii) of obtaining a total of 6?

(iii) of obtaining a total of 10?

(iv) of obtaining the same number on both dice?

(v) of obtaining a total more than 9?

(vi) that the sum of the two numbers appearing on the top of the dice is 13?

(vii) that the sum of the numbers appearing on the top of the dice is less than or equal to 12?

(viii) that the product of numbers appearing on the top of the dice is less than 9.               [CBSE 2014]

(ix) that the difference of the numbers appearing on the top of two dice is 2.                     [CBSE 2014]

Solution:

GIVEN: A pair of dice is thrown

TO FIND: Probability of the following:

Let us first write the all possible events that can occur

(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),

Hence total number of events is 

(i) Favorable events i.e. getting the sum of numbers on the dice equal to 8

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

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 8 is 5

We know that PROBABILITY = 

Hence probability of getting the sum of numbers on the dice equal to 8 

(ii) Favorable events i.e. getting the sum of numbers on the dice equal to 6

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

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 6 is 5

We know that PROBABILITY = 

Hence probability of getting the sum of numbers on the dice equal to 6 is 

(iii) Favorable events i.e. getting the sum of numbers on the dice equal to10 is (4, 6), (5, 5) and (6, 4)

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 6 is 3

We know that PROBABILITY = 

Hence probability of getting the sum of numbers on the dice equal to10 is 

(iv) Favorable events i.e. getting the same number on both the dice

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

Hence total number of favorable events i.e. the same number on both the dice is 6

We know that PROBABILITY = 

Hence probability of getting the same number on both the dice= 

(v) Favorable events i.e. getting the sum of numbers on the dice is greater than 10

is (5, 5), (5, 6), (6, 4), (6, 5) and (6, 6)

Hence total number of favorable events i.e. getting the total of numbers on the dice greater than 9 is 6

We know that PROBABILITY = 

Hence probability of getting the total of numbers on the dice greater than 9 is 

(vi) Favorable events i.e. getting the sum of both numbers appearing on the top of the dice is 13 is 0 since the highest sum of score we can get is 12

Hence probability of getting the sum of both numbers appearing on the top of the dice 13 is equal to

(vii) Favorable events i.e. getting the sum of both numbers appearing on the top of the dice less than or equal to 12 is a sure event

Hence probability of getting the sum of both numbers appearing on the top of the dice less than or equal to 12 is equal to

(vii) Favourable outcomes for getting the product of numbers less than 9 are

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

Thus, the number of favourable outcomes are 16.

$\therefore \mathrm{P}($ getting the product of numbers less than 9$)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{16}{36}=\frac{4}{9}$


(vii) Favourable outcomes for getting the difference of the numbers as 2 are

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

Thus, the number of favourable outcomes are 8.

$\therefore P($ getting the difference of the numbers as 2$)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{8}{36}=\frac{2}{9}$

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