Question

a die is loaded so that probability of an even number is twice the probability of odd number given number are equally likely as well as odd numbers find the probability that an even even number appears uppermost prime number appears uppermost an odd number appears uppermost a perfect square appears uppermost​

Answer 1

Answer:

4/9

Step-by-step explanation:

Number of dice = 1 (Given)

Total number of faces = 6 (Given)

A die is rolled such that each even number occurs twice and odd number occurs once.

Therefore the total possible sample space will be = {1,1,2,3,3,4,5,5,6}

Thus n(s)=9

P(Greater than 3) to occur.

Hence, the number of favorable outcomes n(E)={4,5,5,6}

= 4

Required probability = n(e)/n(s)

= 4/9

Answer 2

Answer:

Step-by-step explanation:

Number of dice = 1 (Given)

Total number of faces = 6 (Given)

A die is rolled such that each even number occurs twice and odd number occurs once.

Therefore the total possible sample space will be = {1,1,2,3,3,4,5,5,6}

Thus n(s)=9

P(Greater than 3) to occur.

Hence, the number of favorable outcomes n(E)={4,5,5,6}

= 4

Required probability = n(e)/n(s)

= 4/9