Problem Solving

-- 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

anju 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

My answer is 7/8.

7 digits, last 3 are 1, 0 is not included.

Therefore we only have to consider 4 digits.

each digit can be filled with 8 letters (2,3,4,5,6,7,8,9)

4 prime numbers and 4 non prime numbers

probability of exactly 1 prime

probability of picking non prime letter for first digit is 1/2
probability of picking non prime letter for second digit is 1/2
probability of picking non prime letter for third digit is 1/2
probability of picking prime letter for fourth digit is 1/2

1/2*1/2*1/2*1/2 = 1/16

Similarly, probability of exactly 0 prime

will be 1/2*1/2*1/2*1/2 = 1/16

1-1/16+1/16 = probability of at least 2 prime


1/16 + 1/16 = 2/16

1-2/16 = 14/16 = 7/8

I dont know where I am going wrong. Whats the source.

parallel_chase wrote:

anju 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

My answer is 7/8.

7 digits, last 3 are 1, 0 is not included.

Therefore we only have to consider 4 digits.

each digit can be filled with 8 letters (2,3,4,5,6,7,8,9)

4 prime numbers and 4 non prime numbers

probability of exactly 1 prime

probability of picking non prime letter for first digit is 1/2
probability of picking non prime letter for second digit is 1/2
probability of picking non prime letter for third digit is 1/2
probability of picking prime letter for fourth digit is 1/2

1/2*1/2*1/2*1/2 = 1/16

Similarly, probability of exactly 0 prime

will be 1/2*1/2*1/2*1/2 = 1/16

1-1/16+1/16 = probability of at least 2 prime


1/16 + 1/16 = 2/16

1-2/16 = 14/16 = 7/8

I dont know where I am going wrong. Whats the source.

You almost got it. There are 4 ways in which one out of 4 numbers is a prime. So you need to multiply 1/16 with 4
probability one prime + probability of no primes = 4 * 1/16 + 1/16 = 5/16

probability of atleast two primes = 1 - 5/16 = 11/16 (B)

You almost got it. There are 4 ways in which one out of 4 numbers is a prime. So you need to multiply 1/16 with 4
probability one prime + probability of no primes = 4 * 1/16 + 1/16 = 5/16

probability of atleast two primes = 1 - 5/16 = 11/16 (B)

Thanks !!! I feel like i just missed it.

Guys,
Do it this simple and standard way:

Probability that a position is prime: 4/8 = 1/2
Probability that a position is not prime: 4/8 = 1/2

P(2 prime) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16 and then permute below
..

P(2 prime) = 1/16 (4c2) = 6/16
Need to permute P1P2## because this can be permuted 6 ways.

P(3 prime) = 1/16 (4c3) = 4/16

P(4 prime) = 1/16 (4c4) = 1/16

Sum the above 6/16 + 4/16 + 1/16 and whola 11/16

A difficult question actually

anju 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

- - - - - - - 7 places . Last 3 was occupied by 1.
Rest 4 will be occupied by:
2 prime, 2 non prime ---- 4C2*4C2
3 prime , 1 non prime ---- 4C3*4C1
4 prime ---- 4C4
No of telephone numbers with at least 2 primes = 4!(4C2*4C2 + 4C3*4C1 +4C4) = 4!*53
Probability = 4!*53 / 4!*8C4
=53/70.
I dont know whether this approach is right or wrong. Please help.

parallel_chase wrote:

anju 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

My answer is 7/8.

7 digits, last 3 are 1, 0 is not included.

Therefore we only have to consider 4 digits.

each digit can be filled with 8 letters (2,3,4,5,6,7,8,9)

Maybe I'm missing something - why don't we also consider 1 as a possible digit for the remaining numbers? The question doesn't say that 1 appears in only the last 3 digits.

So, there should be 9 possibilities for the other 4 spots: {1, 2, 3, 4, 5, 6, 7, 8, 9}, of which 4 {2, 3, 5, 7} are primes.

If we want at least 2 primes, then we want either 2 primes, 3 primes or 4 primes. Alternatively, the only things we do NOT want are 0 primes and 1 prime.

Using the negative approach:

Prob 0 primes = 4C0 * (5/9)^4
Prob 1 prime = 4C1 * (5/9)^3 * (4/9)^1

Prob NOT getting 0 or 1 prime = 1 - (Prob 0 Primes + Prob 1 Prime)

However, we should be able to see right away that the final answer is not going to be a nice neat fraction, which means we won't get any of the listed choices. So, it's apparently correct to ignore the digit "1", although I really can't see why that's the case (unless the question has been incorrectly reproduced).

* * *

So, reading the question as "1 appeared in only the last 3 places", we do indeed have a binary situation - the probability of a prime is 1/2 and the probability of a non-prime is 1/2. Accordingly, this is a great question to apply Pascal's Triangle (discussed in detail here: https://www.beatthegmat.com/coin-flip-qu ... 17911.html)

Looking at the n=4 row (since we have 4 digits to fill), we see:

1 4 6 4 1

We want at least 2 primes, so we add 6 + 4 + 1 to get 11 desired results.
To find the total number of possibilities, we add the entire row to get 16.

Therefore, the probability of getting at least 2 primes out of 4 digits is 11/16.

bourne159 wrote:

parallel_chase wrote:

anju 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

My answer is 7/8.

7 digits, last 3 are 1, 0 is not included.

Therefore we only have to consider 4 digits.

each digit can be filled with 8 letters (2,3,4,5,6,7,8,9)

4 prime numbers and 4 non prime numbers

probability of exactly 1 prime

probability of picking non prime letter for first digit is 1/2
probability of picking non prime letter for second digit is 1/2
probability of picking non prime letter for third digit is 1/2
probability of picking prime letter for fourth digit is 1/2

1/2*1/2*1/2*1/2 = 1/16

Similarly, probability of exactly 0 prime

will be 1/2*1/2*1/2*1/2 = 1/16

1-1/16+1/16 = probability of at least 2 prime


1/16 + 1/16 = 2/16

1-2/16 = 14/16 = 7/8

I dont know where I am going wrong. Whats the source.

You almost got it. There are 4 ways in which one out of 4 numbers is a prime. So you need to multiply 1/16 with 4
probability one prime + probability of no primes = 4 * 1/16 + 1/16 = 5/16

probability of atleast two primes = 1 - 5/16 = 11/16 (B)

Very good thinking. Hats off! 8)

Thanks Stuart for clearing the doubts

I approached with 9 digits, as 1 included.
none of the answers were matching

it should have said "1 appeared in the last 3 places only"

You almost got it. There are 4 ways in which one out of 4 numbers is a prime. So you need to multiply 1/16 with 4
probability one prime + probability of no primes = 4 * 1/16 + 1/16 = 5/16

probability of atleast two primes = 1 - 5/16 = 11/16 (B)


Why do we multiply 1/16 by 4?