1. ## 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.

2. ## Re: Mutually/non mutually events

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?)

3. ## Re: Mutually/non mutually events

Hello, terminator!

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: .$\displaystyle J\heartsuit,\;Q\heartsuit,\;K\heartsuit$

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