However, I've my own approach to solve this problem.
I would go this way -
Let the reqd phone no be _ _ _ _ 111
Although not clearly mentioned in the question, I would ignore the possibility of '1' for the remaining 4 places as John remembers that it appeared in the last 3 places.
Numbers we can include = 2,3,4,5,6,7,8,9
Prime nos: 2,3,5,7
Non-prime: 4,6,8,9
Thus, the probability that the selected no is prime or non-prime is 4/8 = 1/2
The question asks us that the reqd no should have atleast 2 prime digits i.e. it can be 2,3 or 4
i.e. it cannot be 0 or 1
We take the approach: Total Probability - opposite probability = Reqd Probability
Case-1: There are Zero primes => Probability = (1/2)*(1/2)*(1/2)*(1/2) = (1/2)^4 = 1/16
Case-2: There is one prime => Probability = {4C1*(1/2)}*(1/2)*(1/2)*(1/2) = 4/16
Adding the above cases, we get 1/16 + 4/16 = 5/16
Thus, probability that the phone no contains atleast 2 primes = 1 - 5/16 = 11/16
Thus, ans = (B)
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