A card is taken at random from ordinary pack of cards.It is then replaced.
Another card is now taken at random from the pack of cards.
Work out the probability of the following.
a.both cards are the picture cards(jacks,queens or kings)
b.neither of cards is a diamond.
c.only one of the cards is an even number.
d.only one cards is a red jack.
e.at least one of the cards is a queen.
Can someone please help me..
Since you are returning the card after the first event, you are dealing with independent probabilities
That is, the outcome of the second event does not depend on the outcome of the first.
a) there are 12 picture cards
so prob(event) = (12/52)(12/52) = ....
b) There are 39 non-diamonds, so....
prob(neither a diamond) = ....
c) Two cases:
EO or OE
prob(EO) = (8/52)(8/52)
prob(OE) = .....
add them
d) do this the same way as c)
e) prob(at least one queen) = 1 - prob(neither a queen)
= 1 - (48/52)(48/52)
= ....
Hello Reiny, I would like to ask on part C, how do you get 8/52? I'm quite confused about it, because I thought that there were 20 even cards and 20 odd cards?
Sorry, 10 odd and 10 even*
Good for you to notice that, I must have lost my card-sense.
Actually we are both wrong. The number cars run from 2 - 10, and in 4 suits that makes 36.
of which 20 are even and 16 are odds
(count them on your fingers)
so make the necessary adjustments in c)
Sure! I'd be happy to help you work out the probabilities for each of these scenarios. Let's go through them one by one:
a. To find the probability that both cards are picture cards (jacks, queens, or kings), we need to find the probability of drawing a picture card on the first draw, and then multiply it by the probability of drawing another picture card on the second draw.
In an ordinary pack of cards, there are 12 picture cards: 4 jacks, 4 queens, and 4 kings. The total number of cards in the deck is 52.
Probability of drawing a picture card on the first draw: 12/52
Since the card is replaced back into the pack, the probability remains the same for the second draw.
Probability of drawing another picture card on the second draw: 12/52
Now we can multiply these probabilities to get the probability of both cards being picture cards:
(12/52) * (12/52) = 144/2704 = 9/169
So, the probability that both cards are picture cards is 9/169.
b. To find the probability that neither of the cards is a diamond, we need to find the probability of not drawing a diamond on the first draw, and then multiply it by the probability of not drawing a diamond on the second draw.
There are 13 diamonds in a deck of 52 cards.
Probability of not drawing a diamond on the first draw: 1 - (13/52) = 39/52
Probability of not drawing a diamond on the second draw: 39/52
Now we can multiply these probabilities to get the probability of neither card being a diamond:
(39/52) * (39/52) = 1521/2704
So, the probability that neither card is a diamond is 1521/2704.
c. To find the probability that only one of the cards is an even number, we need to consider two cases: one card is even and the other card is not, or one card is not even and the other card is.
There are 20 even-numbered cards in a deck of 52 cards (2, 4, 6, 8, 10 for each suit).
Case 1: One card is even and the other card is not even.
Probability of drawing an even card on the first draw: 20/52
Probability of drawing a non-even card on the second draw: 32/52
Case 2: One card is not even and the other card is even.
Probability of drawing a non-even card on the first draw: 32/52
Probability of drawing an even card on the second draw: 20/52
Now we can add the probabilities of the two cases to get the probability of only one card being an even number:
(20/52) * (32/52) + (32/52) * (20/52) = 640/2704
So, the probability that only one of the cards is an even number is 640/2704.
d. To find the probability that only one card is a red jack, we need to consider two cases: one card is a red jack and the other card is not, or one card is not a red jack and the other card is.
There are 2 red jacks in a deck of 52 cards (jack of hearts and jack of diamonds).
Case 1: One card is a red jack and the other card is not.
Probability of drawing a red jack on the first draw: 2/52
Probability of not drawing a red jack on the second draw: 50/52
Case 2: One card is not a red jack and the other card is.
Probability of not drawing a red jack on the first draw: 50/52
Probability of drawing a red jack on the second draw: 2/52
Now we can add the probabilities of the two cases to get the probability of only one card being a red jack:
(2/52) * (50/52) + (50/52) * (2/52) = 100/2704
So, the probability that only one card is a red jack is 100/2704.
e. To find the probability that at least one of the cards is a queen, we need to consider two cases: one card is a queen and the other card is not, or both cards are queens.
There are 4 queens in a deck of 52 cards.
Case 1: One card is a queen and the other card is not.
Probability of drawing a queen on the first draw: 4/52
Probability of not drawing a queen on the second draw: 48/52
Case 2: Both cards are queens.
Probability of drawing a queen on the first draw: 4/52
Probability of drawing another queen on the second draw: 3/52
Now we can add the probabilities of the two cases to get the probability of at least one card being a queen:
(4/52) * (48/52) + (4/52) * (3/52) = 336/2704
So, the probability that at least one of the cards is a queen is 336/2704.
I hope this helps! Let me know if you have any further questions.