prachich1987 wrote:John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?
a) 15/16
b) 11/16
c) 11/12
d) ½
e) 5/8
Since 1 is in the last 3 places, we have 7-3 = 4 positions that need to be filled.
Since 1 and 0 cannot be used, we have 8 digits from which to choose: 2,3,4,5,6,7,8,9.
Of these 8 digits, half are prime.
So if p=prime and n=not prime, P(p) = 1/2 and P(n) = 1/2.
Let's determine P(exactly 2 prime):
P(ppnn) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16.
To determine P(exactly 2 prime), we multiply by the number of ways to arrange ppnn:
P(exactly 2 prime) = 1/16 * 4!/(2!*2!) = 6/16.
Now here's a neat trick. Let P(0) = no primes, P(1) = exactly 1 prime, P(2) = exactly 2 primes, etc.
P(0) + P(1) + P(2) + P(3) + P(4) = 1.
Probabilities are symmetrical:
P(0) = P(4)
P(1) = P(3)
Thus:
P(4) + P(3) + P(2) + P(3) + P(4) = 1.
Subsituting P(2) = 6/16, we get:
P(4) + P(3) + 6/16 + P(3) + P(4) = 1.
P(3) + P(4) = 5/16
Since P(at least 2 primes) = P(2) + P(3) + P(4), we get:
P(2) + P(3) + P(4) = 6/16 + 5/16 = 11/16.
The correct answer is
B.
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