Question 699206: You have 10 playing cards - the ace (one) to 10 of hearts, inclusive. You choose 3 of these cards at random. What is the probability that the lowest numbered card in these three is the 5?
Not really sure how to start. I have tried different combination. But they don't seem to come to sensible answers.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Consider drawing three cards but doing it one at a time without replacement. This is the same as just drawing three at random.
Since out of the 10 cards, 4 are lower than 5 and 6 are 5 or greater, the probability that the first card is 5 or greater is 
Then, given that the first one drawn is, in fact, 5 or greater, what remains is 9 total cards, 5 of which are 5 or greater. Probability of drawing a second card that is 5 or greater is then  .
And then, given that the first and second draws were 5 or greater, what remains is 8 total cards, 4 of which are 5 or greater. Probability of drawing a third card that is 5 or greater is then  .
Calculated this way, each of the individual probabilities are independent of each other so the total probability is the product:
You can do your own arithmetic.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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