Imagine having a urn with 20 balls, of which 8 are black and the remaining white. You are to draw 5 balls out of the urn. What is the probability that at least three of the balls drawn are white?
prob(white) = 12/20
= 3/5
prob(not white) = 2/5
you want prob that
3 are white or 4 are white or 5 are white
= C(12,3) (3/5)^3 (2/5)^9 + C(12,4) (3/5)^4 (2/5)^8 + C(12,5) (3/5)^5 (2/5)^7
I will do the first term, you do the rest
= .012457 +....
I got .155397 but that seems too little a number considering there are more white than black balls?
Would .155397 be correct?
To calculate the probability of drawing at least three white balls out of five, we need to consider the different scenarios that satisfy this condition.
Scenario 1: Drawing exactly 3 white balls and 2 black balls.
To calculate the probability of this scenario, we use the combination formula. There are 8 white balls in the urn, so we need to choose 3 of them. Similarly, there are 12 black balls, and we need to choose 2 of them. The probability of this scenario is given by:
P(3 white and 2 black) = (C(8,3) * C(12,2)) / C(20,5)
Scenario 2: Drawing exactly 4 white balls and 1 black ball.
Using the combination formula, we can calculate the probability of this scenario:
P(4 white and 1 black) = (C(8,4) * C(12,1)) / C(20,5)
Scenario 3: Drawing all 5 white balls.
The probability of this scenario is simply the combination of choosing all 5 white balls out of the 8 available:
P(5 white) = C(8,5) / C(20,5)
Finally, to calculate the probability of drawing at least three white balls, we sum up the probabilities of all these scenarios:
P(at least 3 white) = P(3 white and 2 black) + P(4 white and 1 black) + P(5 white)
Therefore, the final probability can be calculated by adding the probabilities of the three scenarios:
P(at least 3 white) = [(C(8,3) * C(12,2)) + (C(8,4) * C(12,1)) + C(8,5)] / C(20,5)
After evaluating the combinations and dividing, we can find the numerical value of the probability.