An urn initially contains 5 white and 7 black balls. Each time a ball is selected, it’s color is noted and
it is replaced in the urn along with two other balls of the same color. Compute the probability that
(a) one black and one white is selected.
(b) exactly two black balls are selected.
a) two cases: bw or wb
case 1:
first ball is b, prob(b) = 7/12
so 3 black balls are added,
now we have 5 w, 10 b
prob(w) = 5/15 = 1/3
prob(case1) = (7/12)(1/3) = 7/36
case 2:
first ball is w, prob(w) = 5/12
so 3 white balls are added,
now we have 8w, 7b
prob(b) = 7/15
prob(case2) = (5/12)(7/15) = 7/36
prob(black and white) = 7/36 + 7/36
= 7/18
b) using the same reasoning:
prob(2black) = (7/12)(10/15) = 7/18
check:
all cases:
ww ---> (5/12)(8/15) = 2/9
wb ---> (5/12)(7/15) = 7/36
bw ---> (7/12)(5/15) = 7/36
bb ---> (7/12)(10/15) = 7/18
add them up:
2/9 + 7/36+7/36+7/18 = 1
case 1
To compute the probabilities, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
Let's start with part (a):
(a) One black and one white ball are selected.
To calculate this probability, we need to consider two cases:
1. Selecting a black ball first and a white ball second.
2. Selecting a white ball first and a black ball second.
Case 1: Selecting black first and white second
The probability of selecting a black ball first is 7 / (5 + 7) = 7/12.
After selecting a black ball, the urn contains 7 black balls and 9 white balls (including the selected black ball).
So, the probability of selecting a white ball second is 9 / (7 + 9) = 9/16.
Therefore, the probability of selecting a black ball first and a white ball second is (7/12) * (9/16).
Case 2: Selecting white first and black second
The probability of selecting a white ball first is 5 / (5 + 7) = 5/12.
After selecting a white ball, the urn contains 5 white balls and 9 black balls (including the selected white ball).
So, the probability of selecting a black ball second is 9 / (5 + 9) = 9/14.
Therefore, the probability of selecting a white ball first and a black ball second is (5/12) * (9/14).
Adding the probabilities from both cases gives us the total probability of selecting one black and one white ball:
(7/12) * (9/16) + (5/12) * (9/14) = 63/192 + 45/168 = 315/672 or approximately 0.46875.
Thus, the probability of selecting one black and one white ball is 315/672 or approximately 0.46875.
Now let's move on to part (b):
(b) Exactly two black balls are selected.
To calculate this probability, we need to consider the case of selecting two black balls.
The probability of selecting a black ball first is 7 / (5 + 7) = 7/12.
After selecting a black ball, the urn contains 7 black balls and 9 white balls (including the selected black ball).
So, the probability of selecting a black ball second is 7 / (7 + 9) = 7/16.
Therefore, the probability of selecting two black balls is (7/12) * (7/16) = 49/192.
Thus, the probability of exactly two black balls being selected is 49/192.
Therefore, the probabilities are as follows:
(a) Probability of one black and one white ball = 315/672 or approximately 0.46875.
(b) Probability of exactly two black balls = 49/192.
To compute the probability of selecting certain outcomes, we need to consider the total number of possible outcomes and the number of favorable outcomes.
(a) Probability of selecting one black and one white ball:
- Let's consider the possible ways of selecting one black and one white ball. We can either select a black ball first and then a white ball, or select a white ball first and then a black ball.
- The probability of selecting a black ball first is given by (7/12) since there are initially 12 balls in the urn (7 black + 5 white) and 7 black balls.
- After selecting a black ball, the number of black balls becomes (7+2=9) in the urn, and the number of white balls remains the same, which is 5.
- Considering that a black ball has been selected and replaced along with two additional balls of the same color, the total number of balls in the urn becomes (12+2=14).
- The probability of selecting a white ball from this urn is given by (5/14) since there are 5 white balls and 14 total balls.
- Overall, the probability of selecting one black and one white ball is (7/12) * (5/14) = 35/168 ≈ 0.2083.
(b) Probability of exactly two black balls being selected:
- To determine the probability of selecting exactly two black balls, we need to consider two scenarios:
1. Selecting a black ball first and a black ball second.
2. Selecting a black ball first, a white ball second, and then a black ball as the third selection.
- The probability of selecting a black ball first is (7/12) as discussed earlier.
- After selecting a black ball, the number of black balls in the urn becomes (7+2=9), and the number of white balls remains 5.
- The probability of selecting a black ball second is (9/14) since there are 9 black balls and a total of 14 balls in the urn after the first selection.
- The overall probability of selecting two black balls in this scenario is (7/12) * (9/14) = 63/168 ≈ 0.375.
- For the second scenario, the probability of selecting a black ball first is (7/12), and the probability of selecting a white ball second is (5/14) as discussed earlier.
- After the second selection, the number of black balls becomes (7+2=9), and the number of white balls becomes (5+2=7).
- The probability of selecting a black ball third is (9/16) since after the second selection, there are 9 black balls and a total of 16 balls in the urn.
- The overall probability of selecting two black balls in this scenario is (7/12) * (5/14) * (9/16) = 315/2688 ≈ 0.1172.
- Adding the probabilities of both scenarios, the probability of exactly two black balls being selected is 63/168 + 315/2688 = 651/2688 ≈ 0.2423.