This is a law of addition probability which states that the probability of A or B equals the probability of A plus the probability of B minus the probability of A and B. Written in mathematical terms, the equation is: P(AorB) = P(A) + P(B) - P(AnB) where P(AnB) = 0 (since you can not pull out a green and black ball at the same time). Let P(A) = Probability of drawing the green ball & let P(B) = Probability of drawing the black ball. Total outcomes is 17. So, P(A) = 4/17 & P(B) = 6/17. Therefore P(green or black) = 4/17 + 6/17 = 10/17.
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 of picking a black ace in one random draw from a normal pack of playing cards is 1/26.
The probability is: 55/200 or 11/40
11/18 x 10/17 = .359
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.
(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.
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.
The probability is: 55/200 or 11/40
11/18 x 10/17 = .359
3/12*3/11 = 9/132, or 6.818%.
2
If you pick enough cards, without replacement, the probability is 1. The probability for a single random draw is 1/26.
It is (7+4)/(7+5+4+4) = 11/20
Suppose there are only 20 cars in the world: 5 will be black and 2 will be black SUVs, so if you pick a black car at random there are 2 chances out of 5 that it will be an SUV, ie probability of 0.4 or 40%