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four digit safe code

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by mals24 » Mon Nov 24, 2008 8:06 am
Probability that code has at least 1 even digit = 1 - Prob. code does not have any even digit.

Numbers available = 0,2,3,5,6,7,8,9
Total number of odd numbers = 4 (3,5,7,9)

Now since the question does not mention that all the digits in the code are different, we can repeat the numbers.

There are 4 options for the 1st digit.
4 options for the 2nd
4 for the 3rd
4 for the fourth.

Number of ways of selecting odd digits = 4*4*4*4
Number of ways of selecting 4 digits = 8*8*8*8

Total probability that all digits are odd = 4^4/8^4 = 1/16

Hence prob. that at least 1 digit is even = 1-1/16 = 15/16
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by moadhia » Mon Nov 24, 2008 8:06 am
The 4 digit code looks like this

_ _ _ _


The question is to find the probability of atleast one even digit, the easiest way to do this would be to find the probability of having no even digit and subtracting it by 1

each digit can have 8 different values so denominator = 8 * 8 * 8 *8 (1)

for all the digits to have only odd numbers the # of ways = 4 * 4 * 4 * 4 (2)

prob of having no even digit = 1 / 16 ( Divide (2) by (1) )


1 - 1/16 = 15/16

Hope this helps.
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Re: four digit safe code

by Stuart@KaplanGMAT » Mon Nov 24, 2008 10:35 am
gmataug08 wrote: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?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16
Understanding probability should make this a 15 second question with a 99% confidence level.

Let's use some common sense. The 8 remaining digits are evenly distributed between even/odd. Right away we know that there will be more than a 50% chance of at least one even digit: eliminate a, b and e.

We know there are 4 digits, so the chance of getting at least one even should be extremely high: out of c and d, d is clearly a better choice.

If you want to actually solve, the solutions posted above are both correct. We could also recognize that since there's a 50% chance that each digit is even (4 evens, 4 odds), this is really a coin flip question, as follows:
If you flip a fair coin 4 times, what's the probability of getting at least 1 head?
Since the only results we do NOT want is 0 heads, we use the "1 minus" approach:

Prob(want) = 1 - Prob(don't want)

Here's the generic coin flip formula:

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

With k=0 and n=4, we get:

4C0/2^4 = 1/16

(some good combinations shortcuts to memorize:
nC0 = 1
nC1 = n
nCn = 1)

So:

Prob(at least 1H) = 1 - 1/16 = 15/16
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