Question

pair of dice is thrown once. Find the probability of getting a total of : (i) At least 9. (ii) Doublet of number (iii) Doublet of prime number.​

Answer 1

Step-by-step explanation:

Given :-

Pair of dice is thrown once.

To find :-

Find the probability of getting a total of :

(i) At least 9.

(ii) Doublet of number

(iii) Doublet of prime number.

Solution :-

Given that

Number of dice is thrown once = 2

We know that

If 'n' dice is thrown once then the total possible outcomes = 6^n

We have,

The total number of all possible outcomes

= 6² = 36

They 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)

i) Probability of getting a total at least 9:-

Total at least 9 means the total must be 9,10,11,12,...

The favourable outcomes = (4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)

Total number of favourable outcomes = 9

We know that

Probability of an event P(E) = Number of favourable outcomes/ Total number of all possible outcomes

Probability of getting a total at least 9

=> P(at least 9) = 9/36 = 1/4

ii) Probability of getting Doublet of number :-

Favourable outcomes for doublet number

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

Total number of favourable outcomes = 6

We know that

Probability of an event P(E) = Number of favourable outcomes/ Total number of all possible outcomes

Probability of getting a doublet number

=> P(Doublet) = 6/36 = 1/6

(iii) Probability of getting Doublet of prime number:-

Favourable outcomes for doublet primes = (2,2),(3,3),(5,5)

Total number of favourable outcomes = 3

We know that

Probability of an event P(E) = Number of favourable outcomes/ Total number of all possible outcomes

Probability of getting a doublet prime number

=> P(Doublet prime) = 3/36 = 1/12

Answer:-

i) Probability of getting a total at least 9 = 1/4

ii) Probability of getting a doublet number = 1/6

iii) Probability of getting a doublet prime number = 1/12

Used formulae:-

→ Probability of an event P(E) = Number of favourable outcomes/ Total number of all possible outcomes

→ If 'n' dice is thrown once then the total possible outcomes = 6^n