newton9 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

Are you sure you've posted the question correctly? For example, perhaps it should read "the digit 1 appeared in only the last 3 places"?

I ask because if we're only worried about the first 4 digits, then when we start to apply the probability formula we get:

Prob = (# desired outcomes)/(total # of possibilities)

and, since we can't use 0, there are 9 digits available for each of the first 4 numbers. Accordingly, the denominator will be:

9*9*9*9

and, no matter how much we cancel out with the numerator, we'll always end up with an odd denominator. Since none of the choices have an odd denominator, there's no choice that's possibly correct.

On the other hand, if we can't use 0 or 1 in the first four numbers, then we have:

8*8*8*8

as our denominator, which could potentially cancel down to match one of the answers.

Let's assume that we can't use 1 and see what happens!

The primes from 2 to 9 are 2, 3, 5 and 7. The non-primes are 4, 6, 8 and 9. Since there are an equal number of each, and since we can use each digit as many times as we want, what we really have is a pseudo-coin flip question. (50/50 chance of each outcome, "with replacement", can always be thought of in terms of coin flips).

Thinking in terms of coin flips, this question is now really asking:

If you flip a fair coin 4 times, what's the probability of getting at least 2 heads?

We want the probability of at least 2 heads, i.e. we want 2, 3 or 4 heads. We could calculate that out, OR we could find the probability of 0 or 1 heads and subtract that from 1; on the GMAT, using the "1 minus" approach is often a much quicker way to attack probability questions.

So, let's use:

Prob (2 or more heads) = 1 - Prob (0 or 1 head)

Now we can simply plug in the coin flip formula:

Prob(k results on n flips) = nCk/2^n

So, Prob(0H or 1H) = 4C0/2^4 + 4C1/2^4 = 1/16 + 4/16 = 5/16

Final step: we want the probability of that NOT happening, so:

1 - 5/16 = 11/16... choose B!

For more on pseudo-coin flip (and coin flip) questions, check out:

https://www.beatthegmat.com/coin-flip-qu ... html#75414

Stuart,

Thanks for the detailed reply.

I got this question from one of the other GMAT forums. So not sure about the veracity of it.

Stuart I did go through your links and you have some amazing posts .Very beneficial indeed. But you say there are pseudo coin flip questions and then there are coin flip questions. Could you explain me the difference. You said pseudo has a prob of 1/2. But all coin questions have a prob of 1/2

Please clarify

E

eaakbari wrote:Stuart I did go through your links and you have some amazing posts .Very beneficial indeed. But you say there are pseudo coin flip questions and then there are coin flip questions. Could you explain me the difference. You said pseudo has a prob of 1/2. But all coin questions have a prob of 1/2

Please clarify

E

Hi,

"pseudo" just indicates that it's techincally not a coin flip problem (since we're not actually flipping coins), but we treat it identical to a coin flip problem.

It's just a label to help us recognize when a problem involving something other than coins can be treated as if it were a coin flip problem.