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)
= ....
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..
4 answers
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)
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)
2 answers