Question 143863: You have a three card deck containing a king, a queen , and a jack. You draw a random card, then put it back and draw a second random card.
calculate the probability that you draw exactly one jack.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! You have a three card deck containing a king, a queen , and a jack. You draw a random card, then put it back and draw a second random card.
calculate the probability that you draw exactly one jack.
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Since you replaced the 1st jack the draws are independent.
P(jack, no jack) = P(jack)P(no jack) = 1/3*2/3 = 2/9
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But [jack/no jack] and [no jack/jack] are mutually exclusive events,
So probability of exactly one jack is 2/9 + 2/9 = 4/9
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Why?
P(no jacks in two draws) = (2/3)^2 = 4/9
P(two jacks in two draws) = (1/3)^2 = 1/9
So, prop(exactly one jack in two draws) = 1-4/9-1/9 = 4/9
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Cheers,
Stan H.
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