ANSWER:

• The probability of getting a twin prime numbered card = 1 / 20 (or) 0.05.

GIVEN:

• A box contains prime numbered cards between 1 to 100.

TO FIND:

• The probability of getting a twin prime numbered card.

EXPLANATION:

Sample space (S) = {1, 2, 3, 4, ……… , 100}

Let A be the event of getting a twin prime number.

Prime numbers = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97)

Twin prime numbers:

• Twin prime numbers are those prime numbers which have a difference of 2.

A = {13, 31, 53, 79, 97}

Difference between the numbers is two.

3 - 1 = 3 - 1 = 5 - 3 = 7 - 9 = 7 - 9 = 2

P(A) = n(A) / n(S)

n(A) = 5

n(S) = 100

p(A) = 5 / 100

p(A) = 1 / 20 (or) 0.05

Hence the probability of getting a twin prime numbered card = 1 / 20 (or) 0.05.

SOME MORE POINTS:

• Probability of a sure event is 1.

• Probability of an IMPOSSIBLE event is 0.

• Probabilty of any event always lies between 0 and 1 (0 ≤ p ≤ 1).

• In a sure event n(A) will be equal to n(S).

• In an impossible event n(A) will be equal to zero.

• P(S) = 1 (always) because here also n(A) = n(S), P(S) = n(S) / n(S) = 1.