It is (7+4)/(7+5+4+4) = 11/20
The probability of getting on the first draw a black ball is: P(B1) = 3/7. The probability of getting a red ball given the event of drawing a black ball on the first draw is: P(R2│B1) = 4/6. The probability of drawing a black ball on the first draw and a red ball on the second draw is: P(B1UR2) = P(B1)∙P(R 2│B1) = (3/7)∙(4/6) = 0.2857... ~ 0.286 ~ ~ 28.6%
5/6
The probability, if the cards are not replaced, is 2,535,000/6,497,400 = 0.3902 approx.
17
1/3
(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
x X over 2 = 2
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.
1 in 52
Two out of nine. 2:9.
The probability of picking a black ace in one random draw from a normal pack of playing cards is 1/26.
The probability is zero, because there are no red balls in the bag.
80% chance, Or 40/50
The probability is: 55/200 or 11/40