Question

A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear.

Answer 1

When a die is thrown, the sample space S is given by, S={ 1,2,3,4,5,6 }.

So, the number of elements in S is, n( S )=6

(i)

Let, A be the event “a prime number appears”.

A={ 2,3,5 }

Thus, n( A )=3

Probability when a die is thrown, a prime number appears is given by,

P( A )= number of outcomes favourable to A Total number of outcomes = n( A ) n( S ) = 3 6 = 1 2

Thus, the probability that when a die is thrown, a prime number appears is 1 2 .

(ii)

Let, A be the event “a number greater than or equal to 3 appears”.

A={ 3,4,5,6 }

Thus, n( A )=4

Probability that when a die is thrown, a number greater than or equal to 3appears is given by,

P( A )= number of outcomes favourable to A Total number of outcomes = n( A ) n( S ) = 4 6 = 2 3

Thus, the probability that when a die is thrown, a number greater than or equal to 3 appears is 2 3 .

(iii)

Let A be the event, “a number less than or equal to 1 appears”.

A={ 1 }

Thus, n( A )=1

Probability that when a die is thrown, a number less than or equal to 1 appears is given by,

P( A )= number of outcomes favourable to A Total number of outcomes = n( A ) n( S ) = 1 6

Thus, the probability that when a die is thrown, a number less than or equal to 1 appears is 1 6 .

(iv)

Let A be the event, “a number more than 6 appears”.

A=ϕ

Thus, n( A )=0

Probability that when a die is thrown, a number more than 6 appears is given by,

P( A )= number of outcomes favourable to A Total number of outcomes = n( A ) n( S ) = 0 6 =0

Thus, the probability that when a die is thrown, a number more than 6 appears is 0.

(v)

Let A be the event, “a number less than 6 appears”.

A={ 1,2,3,4,5 }

Thus, n( A )=5

Probability that when a die is thrown, a number less than 6 appears is given by,

P( A )= number of outcomes favourable to A Total number of outcomes = n( A ) n( S ) = 5 6

Thus, the probability that when a die is thrown, a number less than 6 appears is 5 6 .