I'll answer the second question first:
Whenever you draw a card, there are two options: it is either red or
black. The "sample space" of an experiment is a list of all possible
outcomes. Four cards with two options each will result in a sample
space of 2*2*2*2 = 16 different combinations. Since this is so few,
I've listed them below:
RRRR
RRRB
RRBR
RBRR
BRRR
RRBB
RBRB
RBBR
BRRB
BRBR
BBRR
BBBR
BBRB
BRBB
RBBB
BBBB
You are only interested in the last five combinations--I am not sure
if you want the probability of drawing exactly 3 black cards (in which
case BBBB is excluded), or the probability of at least 3 black cards
(in which case BBBB is included). That can easily be accounted for in
the calculation.
Let's look at the combination BBRB. First we must draw one black card
from a deck of 52--which contains 26 black cards. So the probability
of the first card being black is 26/52 or 0.500. The second card must
also be black, but since we have already removed a card from the deck,
the probability is no longer 0.500; instead, we now select amongst the
remaining 25 black cards in the deck, which now consists of 51 cards.
So the probability of the second black card is 25/51 = 0.490. Next we
wish to draw a red card. Since no red cards have yet been chosen,
there are 26 available reds in a deck of size 50, yielding a
probability of 26/50 = 0.520. Finally, the last black card will be
selected from 24 blacks in a deck of 49 cards, for a probability of
24/49 = 0.490 (I am rounding to three decimal places on all
calculations). The total probability of selecting BBRB is found by
multiplying the individual card probabilities (multiplication is the
statistician's way of saying "and"; we want a black and a black and a
red and a black, so we multiply). 0.500 * 0.490 * 0.520 * 0.490 =
0.0624.
The statistician's way of saying "or" is addition. So to find the
probability of BBBR or BBRB, we simply add the probabilities of each
of the combinations. Below, I have the work shown (I use fractions
until the end in order to preserve accuracy):
BBBR: (26/52)*(25/51)*(24/50)*(26/49) = 52/833 = 0.62425
BBRB: (26/52)*(25/51)*(26/50)*(24/49) = 52/833
BRBB: (26/52)*(26/51)*(25/50)*(24/49) = 52/833
RBBB: (26/52)*(26/51)*(25/50)*(24/49) = 52/833
Total probability of selecting exactly 3 black cards = 4*(52/833) =
0.24970
If you want the probability of at least four black cards, you need to
add the probability of BBBB:
BBBB: (26/52)*(25/51)*(24/50)*(23/49) = 46/833 = 0.055222
Total probability of at least four black cards: 0.30492
For the first question, it does not really matter that you are drawing
five cards, since the event you are concerned about takes place on the
third draw. You can lay out a sample space for the first three cards
like we did above. For 3 cards, there should be 2*2*2 = 8
combinations:
RRR
RRB
RBR
BRR
BBR
BRB
RBB
BBB
Of these, four are of importance in solving your problem. We can
approach this in a similar way as the other problem:
RRB: (26/52)*(25/51)*(26/50) = 13/102
RBR: (26/52)*(26/51)*(25/50) = 13/102
RBB: (26/52)*(26/51)*(25/50) = 13/102
BBB: (26/52)*(25/51)*(24/50) = 12/102
Total probability of a black on the third draw = 3*(13/102) + (12/102)
= 0.500, probably as you expected.
I hope this is clear to you. Thanks for the question.
xargon-ga |