Asked by m
The numbers 1-10 are placed on cards, and the cards are put in a hat. If two cards are drawn, in order, without replacement, what is the probability that one number is even and the other a prime?
I know there are 4 primes-2,3,5,7 and 5 even numbers 2,4,6,8,10
not sure how to figure the probability
I know there are 4 primes-2,3,5,7 and 5 even numbers 2,4,6,8,10
not sure how to figure the probability
Answers
Answered by
Reiny
It could be
PE or EP
what messes things up is that the 2 is in both groups
PE:
case 1: the 2 was picked as the prime
prob of that is (1/10)(4/9)
case 2: the 2 was not picked as the prime
prob of that is (3/10)(5/9)
so prob (PE) = 4/90 + 15/90 = 19/90
EP:
case 1: the 2 was picked as the even
prob = (1/10)(3/9) = 3/90
case 2: the 2 was not picked as the even
prob = (4/10)(4/9) = 16/90
prob(EP) = 3/90 + 16/90 = 19/90
so prob of (one prime and one even) = 19/90 + 19/90 = 19/45
PE or EP
what messes things up is that the 2 is in both groups
PE:
case 1: the 2 was picked as the prime
prob of that is (1/10)(4/9)
case 2: the 2 was not picked as the prime
prob of that is (3/10)(5/9)
so prob (PE) = 4/90 + 15/90 = 19/90
EP:
case 1: the 2 was picked as the even
prob = (1/10)(3/9) = 3/90
case 2: the 2 was not picked as the even
prob = (4/10)(4/9) = 16/90
prob(EP) = 3/90 + 16/90 = 19/90
so prob of (one prime and one even) = 19/90 + 19/90 = 19/45
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