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i will try to solve, but not strong in this topic
we left with 4 odd digits (3,5,7,9), without any restrictions the first place of the cod can be filled with any of the 4 digis, the same is true for the 2-place, 3, and 4
so there are 4*4*4*4=4^4 outcomes to have code with only odd digits (exluding 1)
the total number of combinations is 8*8*8*8=8^4, as we have 0.2.3.5.6.7.8.9 to chose from
(4^4)/(8^4)=1/16 it is prob that code will consists only odd digits, to answer the problem we must substract 1/16 from 1
1-1/16=15/16

Anytime you have a problem like this one with the word ATLEAST in it, remember one thing: calculate the probability of it NOT happening at ALL and subtract it from 1 which will give you the probability of it happening once.

So lets do this one:

4 digit code: _ _ _ _ possible values (0,2,3,5,6,7,8,9)

so (1/2)^4 = 1/16. You said you got this far. Now all you have to do is subtract it from 1


(In this question, the probability of an even number for EACH of the digits is 4/8 or 1/2. Therefore, the probability of it NOT being an even number is the same: 1/2. So we will use the same number. If the numbers were different, say the probability of something happening was 5/8 then the probability of it NOT happening would b 3/8.)

1 - 1/16 = 15/16.

There's your answer. I hope that helps!

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I used to get confused about this as well.

Perhaps you can think of it this way:

You're thinking of the probability of getting only one even as this: [Even](1/2)*[Odd](1/2)*[Odd](1/2)*[Odd](1/2), but you have to keep in mind that that is the probability of getting ONLY one even AND that even occurring in the FIRST DIGIT. The probability of only ONE even is actually 1/4 because there are 4 different ways to get ONE even and THREE ODDS. (1/4=4*1/16)

Now you could potentially solve this way.

Let's keep going with your method.

Probability of 2 Evens is 6*1/16=6/16. You can find that there are six ways of getting two evens and two odds by using the formula N!/r!*r2!*....r100 where N is the total number of SLOTS to fill and r is the number of repeating items for the first type of time, r2 is the number of repeating for the second, etc. THis is the same formula you use to see how many ways you can rearrange a group of letters with letters that repeat. And that's essentially what we're finding: how many ways can we arrange the "letters" EEOO. Well there's 4 spaces, E repeats twice as does O. So 4!/(2!*2!)=6.

Probability of 3 evens is the same as 1 even and three odd= 4/16

Probability of 4 evens is 1/16

Then you can add all the probabilities 4/16 + 6/16 + 4/16 + 1/16= 15/16

Obviously this is a little more work but I found for me that understanding how to solve probabilities in more than one way really helps me to understand WHAT I'm actually solving instead of just getting attached to a memorized formula.

Hope this helps.

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Hi:
Given that no's 1 and 4 are out, this only leaves following no's - 0, 2, 3, 5,6,7,8,9

On any given position any of these 8 no's can be present. So total combinations = 8x8x8x8
To find one or more of even no's lets look at possibility where no even nos are present on any location.
So only 3,5,7,9 can be used on any given location giving us total of 4x4x4x4 combos

So the probability of getting non-even no's on all 4 locations is (4x4x4x4)/(8x8x8x8) = 1/16
Means that probability of *not* getting "non-even on all 4 locations" = 1-1/16 = 15/16

but not getting "non-even on all 4 locations" means getting even on one or more locations. Hence probability = 15/16

man... I've got to remember that 0 is an even number!! I was stuck on this problem for 10 minutes, and my first approach was the correct one... I just counted the possibility of an even as 3/8 rather than 1/2 because I neglected that zero!

I have a difficulty in understanding the above.
A safe code consists of 4 digits.
Can 0 be at the first place???
according to me the total no. of combinations should be = 7*8*8*8
Please correct me if I am wrong

Even:0,2,4,6,8
Odd:1,3,5,7,9
Now 1 and 4 are discarded leaving out with 8 digits.
Each digit can be either odd or even so 2^4=16 possibilities(combinations).
Now amongst them one combination has all odd digits(3,5,7,9)
so Required Probability(1-11/6)=15/16

can any of these 8 digits repeat in the 4 digit code or they all distinct?

Can someone post an easier to understand explanation?

since nowhere it is mentioned that the digits can not repeat,u can conclude that they can. )

from the q.stem u see that there are 8 available digits-

0 2 3 5 7 6 8 9 (exclude 1 and 4, because it was stated in the q.stem )

remember -every time u see a q.,which asks u "at least of smth" or "at most of smth", try to find the contrary one,i.e. since we are asked to find out the probability that the code has at least ONE even digit, we need to find out : 1 -"the probability of NO even digit at all"

so, not even digits in our set of digits are 3 5 7 9 (4 digits)

since in total we have 8 digits, so for every 4-digit code we have (4/8)*(4/8)*(4/8)*(4/8)=(1/2)^4=1/16

1-1/16=15/16 ( the probability of at least one even digit)

hope it is clear and I could help u some way

atleast 1 even = 1 - no even.
As explained already P,no even = P , all ODD =>
favourable cases = 3,5,7,9 = 4 ^4
total cases = 8^4
hence P all ODD = (4/8)^4 = 1/16
hence P atleast 1 even = 1-1/16 = 15/16

At least even = all - odds
all combination = 8^4
Odd combination = 4^4
P = 1 - 4^4/8^4 = 1-1/16 = 15/16

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