Question

Two numbers are randomly selected without replacement from the set{1,2,3,4,5}.Find the probability that:

a)both numbers are odd

b)both numbers are prime

Answer 1

Answer to the question)
Given:

Two numbers are selected

and not replaced back

total number of outcomes = 5 {1,2,3,4,5}

.

Part a)

Odd numbers = 3 , {1,3,5}

Number of ways of selecting 2 numbers out of 3 odd numbers = 3C2

Number of ways of selecting 2 numbers out of 5 total numbers = 5C2

P(both number are odd) = number of ways of selecting 2 odd numbers / total number of ways of selecting 2 numbers

P(both numbers are odd) = 3C2 /5C2

[when replacement is not done, one can use the countinng technique of combination]

[formula of combination is : nCr = n! / (n-r)! * r!]

[where r! is called r factorial : r! = r*(r-1)*(r-2)*.....*3*2*1 = product of all natural numbers from 1 to r]

.

Thus we get:

P(both numbers are odd) = [3!/2!*1!] / [5!/2!*3!]

P(both numbers are odd) = 3 / 10

P(both numbers are odd) = 0.30

.

Part b)

Total number of prime numbers is : {2,3,5 } , so 3 numbers are prime

Number of ways of selecting 2 numbers out of 3 prime numbers = 3C2

Number of ways of selecting 2 numbers out of 5 total numbers = 5C2

P(both number are odd) = number of ways of selecting 2 prime numbers / total number of ways of selecting 2 numbers

Thus we get:

P(both numbers are prime) = 3C2 /5C2

P(both numbers are prime) = 3/10

P(both numbers are prime) = 0.30