From a box containing 4 white, 5 green, and 5 yellow balls, 2 balls are drawn at a time without replacing the first before the second is drawn. Find the probability that 1 white and 1 yellow ball are drawn.
Round your answer to 4 decimal places.
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Total number of possibilities: 4 white + 5 green + 5 yellow = 14 total balls.
Probability of choosing 1 white ball: 4 / 14
Probability of choosing 1 yellow ball given that 1 white ball was already chosen: 5 / 13 (if we don't replace the balls, there will be 13 left after 1 white ball is taken out)
Probability of choosing 1 white then 1 yellow: (4 / 14) x (5 / 13) = .1099
this is a conditional probability problem its not done that way wish it was that wasy
Conditional probability problem? Of course it is.
P(white AND yellow) = P(white) * P(yellow | white)
The probability of getting one white is 4 / 14.
The probability of getting a yellow ball on the second try is dependent on the selection of a first ball, so this is a conditional probability problem. The probability of getting a yellow ball given that white has occurred (the condition) is 5 / 13.
You multiply the two fractions to get the probability of getting a white and then a yellow ball.
I'm not sure what exactly could be required of you if this doesn't work...