14/27 ~ 0.519 ~ 51.9%
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EXPLANATION
For bag 1: Probability of drawing a black ball; P(B1) =
6/9. A white one; P(W1) =3/9
For bag 2: P(B2) =
5/9. P(W2) =
4/9.
Probability of drawing a black ball from each bag: P(B1UB2) =
(6/9)∙(5/9) =
10/27
Probability of drawing a white ball from each bag: P(W1UW2) =
(3/9)∙(4/9) =
4/27
The probability of drawing 2 balls of the same color is the sum of the above
probabilities:
P(2 same color balls) =
P(B1UB2) + P(W1UW2) =
10/27 +4/27 =
14/27 =
0.5185...
P(2 same color balls) ~ 0.519 ~ 51.9%
2
11/18 x 10/17 = .359
The probability of a black ball in bag 1 is 4/12 = 1/3. If you add 3 black balls to bag 2, it will contain 5 black balls out of 15: the probability of a black ball being 3/15 = 1/3.
The probability is: 55/200 or 11/40
0.8.
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
If you draw enough balls, without replacement, the probability is 1.The answer depends onhow many balls are drawn, andwhether or not they are replaced.Unfortunately, your question gives no information on these matters.
2
11/18 x 10/17 = .359
The probability of a black ball in bag 1 is 4/12 = 1/3. If you add 3 black balls to bag 2, it will contain 5 black balls out of 15: the probability of a black ball being 3/15 = 1/3.
The probability is zero, because there are no red balls in the bag.
The probability is: 55/200 or 11/40
3/12*3/11 = 9/132, or 6.818%.
0.8.
There is no event defined for which a probability can be calculated!
It is (7+4)/(7+5+4+4) = 11/20
with replacement: binominal distribution f(k;n,p) = f(0;5,5/12) without replacement: hypergeometric distribution f(k;N,m,n) = f(0;12,5,5)