Mutually/non mutually events

A card is taken from a deck of cards and then replaced. After the deck is shuffled, a second card is taken from the deck. caluculate the probabilities.

a. Both cards are diamonds or both cards are hearts.

b. Both cards are black or both cards are hearts.

c.Both cars are 2's or both cards are face cards.

d.Both cards are hearts and both cards are face cards.

a. P(A or B) = P(A) + P(B) Exclusive events

P( 1/4 x 1/4) + P(1/4 x 1/4) = 1/8

b. P( A or B) = P(1/2 x 1/2) + P( 1/4 + 1/4) = 5/16

c. P(A or B) = P(1/13 x 1/13) + P(12/52 + 12/52) = 10/169

d. P( A or B) = P(A) + P(B) - P(A or B) non -mutually events

= (1/4 x1/4) + (12/52 x 12/52) - P(1/16 x 36/676) = 0.1124

Need feedback.

Thanks.

Re: Mutually/non mutually events

Quote:

a. P(A or B) = P(A) + P(B) Exclusive events

P( 1/4 x 1/4) + P(1/4 x 1/4) = 1/8

b. P( A or B) = P(1/2 x 1/2) + P( 1/4 **x** 1/4) = 5/16

c. P(A or B) = P(1/13 x 1/13) + P(12/52 **x** 12/52) = 10/169

d. P( A or B) = P(A) + P(B) - P(A **and ** B) non -mutually events

= (1/4 x1/4) + (12/52 x 12/52) - P(1/16 x 36/676) = 0.1124

NUmerical answers look right to me. minor corrections in red (typing errors?)

Re: Mutually/non mutually events

Hello, terminator!

I don't agree with your answer to (d).

Quote:

A card is taken from a deck of cards and then replaced.

After the deck is shuffled, a second card is taken from the deck.

Calculate the probability:

(d) Both cards are hearts **and** both cards are face cards.

There are three cards which are hearts **and** face cards: .$\displaystyle J\heartsuit,\;Q\heartsuit,\;K\heartsuit$

$\displaystyle P(\text{both heart face cards}) \;=\;\frac{3}{52}\cdot\frac{3}{52} \;=\; \frac{9}{2704}$