Question: Two cards are drawn from a well-shuffled ordinary pack of 52 cards. The first card is replaced before selecting the second card. Find the following probability.

- Both are kings
- First is king and second is queen
- One is King and one is queen
- Both are faced cards
- First is card of diamonds and the second is an ace.

Solution:

__Probability that Both are kings__

A= Event (1^{st} card drawn is king)

B= Event (2^{nd} card drawn is king)

P(A)= Total kings in the deck/ Total cards in the deck

P(A)= 4/52

P(B)= Total kings in the deck/ Total cards in the deck

P(B)= 4/52

P(A and B) = P(A) * P(B)= 4/52* 4/52= 1/169

__Probability that First is king and second is queen__

__ __

A= Event (1^{st} card drawn is king)

B= Event (2^{nd} card drawn is queen)

P(A)= Total kings in the deck/ Total cards in the deck

P(A)= 4/52

P(B)= Total queens in the deck/ Total cards in the deck

P(B)= 4/52

P(A and B) = P(A) * P(B)= 4/52* 4/52= 1/169

__Probability that One is King and one is queen__

A= Event (1^{st} card drawn is king)

B= Event (2^{nd} card drawn is queen)

P(A)= Total kings in the deck/ Total cards in the deck

P(A)= 4/52

P(B)= Total queens in the deck/ Total cards in the deck

P(B)= 4/52

P(A and B) = P(A) * P(B)= 4/52* 4/52= 1/169

__Probability that Both are faced cards__

A= Event (1^{st} card drawn is face card)

B= Event (2^{nd} card drawn is face card)

P(A)= Total face cards in the deck/ Total cards in the deck

P(A)= 12/52

P(B)= Total face cards in the deck/ Total cards in the deck

P(B)= 12/52

P(A and B) = P(A) * P(B)= 12/52* 12/52= 9/169

__Probability that First is card of diamonds and the second is an ace.__

A= Event (1^{st} card drawn is of diamonds)

B= Event (2^{nd} card drawn is of ace)

P(A)= Total diamond cards in the deck/ Total cards in the deck

P(A)= 13/52

P(B)= Total ace cards in the deck/ Total cards in the deck

P(B)= 4/52

P(A and B) = P(A) * P(B)= 13/52* 4/52= 1/52

__Question__**:** An integer is chosen at random from the first 200 positive integers. What is the probability that the integer chosen is divisible by 6 or by 8?

** Solution**:

Let S be the Sample Space; then

S={1,2,3,4,………….,200}, n(S)=200

Let A be the event i.e., the integer divisible by 6

A={6,12,18,……,198} , n(A)=33

P(A)=n(A)/n(S)=33/200 ………………….(i)

Let B is the event i.e., the integer divisible by 8

B={8,16,24,……,192}, n(B)=25

P(B)=n(B)/n(S)=8/200 …………………. (ii)

And (A intersection B) is the event, the integers divisible by both 6 and 8

(A intersection B) = {24,48,72,….,192}, n(A intersection B)=8

P(A intersection B)=n(A intersection B)/n(S)=8/200 ………………. (iii)

According to the formula (Non mutually exclusive events)

P(AUB)=P(A)+P(B)-P(A intersection B)

=33/200+25/200-8/200 = 1/4Ans.