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# Specular problem, but different approach to solve them?

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Intern
Joined: 30 Aug 2010
Posts: 13

Kudos [?]: 44 [0], given: 2

Specular problem, but different approach to solve them? [#permalink]

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21 Sep 2010, 07:30
Hi,
as already said I've a lot problem with probabilities and combinations.
I encountered the following problems, they seem very similar to me. So I tried to solve them with the same method. But it didn't work.

The problem are:
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?
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?

1) I have four slots _ _ _ _ that can contain only {0,2,3,5,6,7,8,9}.
Four of them are even, four are odd.
So the probability of having an even digit is $$4/8=1/2$$, as well having an odd digit.

The solution can be calculated as:
Probability of at least one even digit = Probability of one even digit + Probability of two even digits + Probability of three even digits + Probability of four even digits

P1 = $$1/2$$
P2 = $$1/2^2=1/4$$
P3 = $$1/2^3=1/8$$
P4 = $$1/2^4=1/16$$
P = P1+P2+P3+P4 = $$15/16$$

2) I have the following situation _ _ _ _ 1 1 1
Empty slots can contain only {2,3,4,5,6,7,8,9}
Prime digits are {2,3,5,7}, not prime digits are {4,6,8,9}.
So the probability of having a prime digit is $$4/8=1/2$$, as well having a non prime digit.

I think I have the same situation of the previous problem: filling 4 empty slots with digits (even or prime) that have the same probability of being in the slot ($$1/2$$) or not being ($$1/2$$).

So I created the solution with the same pattern of the previous problem:
Probability of at least two prime digits = Probability of two prime digits + Probability of three prime digits + Probability of four prime digits

P2 = $$1/2^2=1/4$$
P3 = $$1/2^3=1/8$$
P4 = $$1/2^4=1/16$$
P = P2+P3+P4 = $$7/16$$

My solution of problem 2 is not correct, while the one of problem 1 is correct. Why?

Kudos and beers for your help!!!

Kudos [?]: 44 [0], given: 2

Retired Moderator
Joined: 02 Sep 2010
Posts: 793

Kudos [?]: 1184 [0], given: 25

Location: London
Re: Specular problem, but different approach to solve them? [#permalink]

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21 Sep 2010, 07:45

I don't think its correct. There is a similar problem plaguing solution 2, which is why you get a wrong answer

And do let me know how you plan to deliver that beer
_________________

Kudos [?]: 1184 [0], given: 25

Intern
Joined: 30 Aug 2010
Posts: 13

Kudos [?]: 44 [0], given: 2

Re: Specular problem, but different approach to solve them? [#permalink]

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21 Sep 2010, 08:07
shrouded1 wrote:

I don't think its correct. There is a similar problem plaguing solution 2, which is why you get a wrong answer

And do let me know how you plan to deliver that beer

Now I got it!
Probability is very tricky to me. I hope that with your advice I can improve in it.

About the beer, pick a London pub. In my next trip to London I will pay a beer for a guy called shrouded1. Then showing your ID to the bartender with shrouded1 as your first name he will give you a free beer.
I'm joking. I appreciate your help!

Kudos [?]: 44 [0], given: 2

Retired Moderator
Joined: 02 Sep 2010
Posts: 793

Kudos [?]: 1184 [0], given: 25

Location: London
Re: Specular problem, but different approach to solve them? [#permalink]

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21 Sep 2010, 12:04
haha ... I shall wait

Time to get a fake ID
_________________

Kudos [?]: 1184 [0], given: 25

Ms. Big Fat Panda
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1915

Kudos [?]: 2203 [0], given: 210

Concentration: General Management, Nonprofit
Re: Specular problem, but different approach to solve them? [#permalink]

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21 Sep 2010, 12:12
LOL. I was in London recently; beers are so overrated.

Kudos [?]: 2203 [0], given: 210

Re: Specular problem, but different approach to solve them?   [#permalink] 21 Sep 2010, 12:12
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