Q4
4. 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
a) 15/16
b) 11/16
c) 11/12
d) ½
e) 5/8
posted by Ramkumar @ 9:12 PM
2 Comments:
At 7:33 AM ,
kart said...
11/16
At 1:20 AM ,
Ramkumar said...
The best answer is B.
Since 1 appears exactly three times, we can solve for the other four digits only. For every digit we can choose out of 8 digits only (without 1 and 0). Since we have 4 prime digits (2, 3, 5, 7) and 4 non-prime digits (4, 6, 8, 9), the probability of choosing a prime digit is ½.
We need at least two prime digits:
One minus (the probability of having no prime digits + having one prime digit):
There are 4 options of one prime digit, each with a probability of (1/2)4.
There is only one option of no prime digit with a probability of (1/2)4.
So: [1- ((1/2)4+(1/2)4*4)] = 11/16.
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