Probability Question?

Question

An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B and so on. There is no replacement of the balls drawn)

Answer 1

Draw 1: 3/10 chance for A
Draw 2: 3/9 * 7/10 chance for B (you need to multiply by 7/10 to account for the chance A drew the red ball on the first draw) = 21/90 = 7/30
Draw 3: 3/8 * 7/10 * 6/9 A = 7/40
Draw 4: 3/7 * 7/10 * 6/9 * 5/8 B = 1/8
Draw 5: 3/6 * 7/10 * 6/9 * 5/8 * 4/7 A = 1/12
Draw 6: 3/5 * 7/10 * 6/9 * 5/8 * 4/7 * 3/6 B = 1/20
Draw 7: 3/4 * 7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/5 A = 1/40
Draw 8: 1 * 7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/5 * 1/4 B = 1/120

probability for A = 3/10 + 7/40 + 1/12 + 1/40 = 36/120 + 21/120 + 10/120 + 3/120 = 70/120 = 7/12
probability for B = 7/30 + 1/8 + 1/20 + 1/120 = 28/120 + 15/120 + 6/120 + 1/120 = 50/120 = 5/12

Answer 2

The formula for the probability of an event happening, P(event) is simply:

P(event) = # favourable outcomes / # of total outcomes possible

e.g. What is the probabilty of flipping heads on a normal coin?

P(heads) = 1/2
there is only 1 favourable outcome (heads) and two possible outcomes (heads or tails)

But your question looks like it isn's so simple, you have several probabilities to figure out. I'll start you out . . .

What is the probability of drawing a red ball on the FIRST try:

P(red) = # favourable outcomes / # of total outcomes possible
P(red) = 3 / 10

BUT lets say that person A DIDN'T get a red ball. If the probability of drawing red is 3/10 then the probability of drawing black must be 7/10

If that happens, you need to figure out the probability of person B drawing a BLACK ball ('cause we want A to get the first red). So:

P(black) = # favourable outcomes / # of total outcomes possible
P(black) = 6 / 9
***remember that one black ball has already been removed***

So we go back to person A. Recalculate the probabilities with 2 black balls missing (there are now 3 red and 5 black).

What is the probability of drawing a red ball on the SECOND try:

P(red) = # favourable outcomes / # of total outcomes possible
P(red) = 3 / 8

BUT we need to factor in the probabilities of A's first try AND B's first try. To find the probabilties of multiple events, we multiply the probabilities of each event.

So basically,
P(A gets a red on his 2nd try) x P(A gets black on first try) x P(B gets black on first try)

7/10 x 6/9 x 3/8 = 21/120

But of course you should calculate probabilities for any more attempts that could happen. By continuing the above steps

Recap:
P(red on 1st try) = 3/10
P(red on 2nd try) = 21/120
etc.
etc.

Answer 3

Sorry, it is not what dato says.

The possible ball sequences are R, BR, BBR, ... and so on up to BBBBBBBR. Each sequence has a probability. You need the sum of the probabilities of the four sequences in which A draws the first red ball. Check that the sum of the probabilities of the other four sequencies brings the total to exactly 1.

For example, the probability of R is 3/10, but also the probability of BBR is (7/10) * (6/9) * (3/8), plus BBBBR and BBBBBBR.

Answer 4

A has 3 in seven chance of getting a red ball. B has 2 in 6 chance of getting a red ball