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probability- safe code

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probability- safe code

by mehrasa » Thu Sep 15, 2011 5:11 am
1) A Four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?
1/4 1/2 3/4 15/16 1/16

[spoiler]OA:D[/spoiler]

2)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?15/16 11/16 11/12 1/2 5/8

[spoiler]OA: B[/spoiler]
Last edited by mehrasa on Thu Sep 15, 2011 6:40 am, edited 1 time in total.
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by Ian Stewart » Thu Sep 15, 2011 6:26 am
mehrasa wrote:1) A Four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?
1/4 1/2 3/4 15/16 1/16
There are 8 digits to choose from (0, 2, 3, 5, 6, 7, 8 and 9), 4 of which are odd. Assuming we pick our codes randomly, and that we can use repeated digits, the probability any individual digit is odd is 4/8 = 1/2. The probability all are odd is thus just (1/2)(1/2)(1/2)(1/2) = 1/16, and the probability at least one digit is even is then 1 - 1/16 = 15/16.
mehrasa wrote: 2)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?
15/16 11/16 11/12 1/2 5/8

[spoiler]OA: B[/spoiler]
The wording of this question is terrible. When it says "the digit 1 appeared in the last three places", this could mean one of two things: that *all* of the last three digits are '1', or that *one* of the last three digits is '1'. The former meaning is what's intended. You are also meant to assume that none of the other digits is '1', something which is completely unclear from the wording of the question. It's one of those practice questions that's more confusing than helpful. In any case, if you have psychic abilities and can work out what the question designer was thinking, you can solve it, as was done in this thread:

https://www.beatthegmat.com/probability-t17876.html
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by shankar.ashwin » Thu Sep 15, 2011 6:34 am
Prob. of atleast one even = 1 - Prob of all odd.

Hence, we find prob of all odd.

Let us consider only the first number for ease of explanation.

The numbers possible are (0,2,3,5,6,7,8,9) - 8 Nos
But we are interested only in picking odd numbers,
So odd No. are (3,4,7,9) - 4 Nos

So prob of choosing odd = 4/8 = 1/2.

Similarly for the 2nd, 3rd and 4th digit of the code we have a 1/2 prob of picking an odd number.

So for all four to be odd, we have (1/2).(1/2).(1/2).(1/2)=1/16.

Now, for atleast one to be even, we have 1 -(1/16) = 15/16.

P.S : The answer choices are giveaways for a educative guess, all other option have very low prob.
mehrasa wrote:1) A Four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?
1/4 1/2 3/4 15/16 1/16

[spoiler]OA:D[/spoiler]
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