A seven-digit combination lock on a safe has zero exactly three times,

Skywalker18 plz explain the formula used.. thanks

If the probability of a certain event is \(p\), then the probability of it occurring \(k\) times in \(n\)-time sequence is: \(P = C^k_n*p^k*(1-p)^{n-k}\)

For example for the case of getting 3 tails in 5 tries:
\(n=5\) (5 tries);
\(k=3\) (we want 3 tail);
\(p=\frac{1}{2}\) (probability of tail is 1/2).

So, \(P = C^k_n*p^k*(1-p)^{n-k}=C^3_5*(\frac{1}{2})^3*(1-\frac{1}{2})^{(5-3)}=C^3_5*(\frac{1}{2})^5\)

OR: probability of scenario t-t-t-h-h is \((\frac{1}{2})^3*(\frac{1}{2})^2\), but t-t-t-h-h can occur in different ways:

t-t-t-h-h - first three tails and fourth and fifth heads;
h-h-t-t-t - first two heads and last three tails;
t-h-h-t-t - first tail, then two heads, then two tails;
...

Certain # of combinations. How many combinations are there? Basically we are looking at # of permutations of five letters t-t-t-h-h, which is \(\frac{5!}{3!2!}\).

Hence \(P=\frac{5!}{3!2!}*(\frac{1}{2})^5\).



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Hope it helps.

Hi, divyadisha u have multipled the probability for 4 digits but number of ways for 3 digits

Hi! Since we are talking about remaining 4 digits. Each digit can either be odd or even (Total ways to select = 2. desired outcome= 1)

so let's say we want to choose first 3 digits as odd digits and last digit as even. The probability will be :-
1/2 *1/2*1/2*1/2= 1/16

But out of remaining 4 digit , any 3 can be odd and one even as there are no constraints in the question. Therefore, total ways of choosing 3 odd digits out of 4 total digits= 4C3

So, the probability = 4*1/16= 1/4

Hello

I am afraid either I do not understand or I am not Ok with your soluce
For me
we can deal with 0 1 2 3 4 5 6 7 8 9
There are 7 places
Let's consider there are still three "0".
_0 _0 _0 _ _ _ _
There are 4 places left
>We know there are no "1".
>So we are left with 2 3 4 5 6 7 8 9 => 8 possibilities for 4 places

Among them we have to chose 3 odd , there are four odds ( 3 5 7 9)
so 4/8*4/8*4/8

And we have to finish with an even number 2 4 6 8 so 4/8

Finally probability is 4/8*4/8*4/8*4/8 = 1/16

I am ok it is only one way to arrange the combination but WE ARE NOT ASKED HOW MANY DIFFERENTS COMBINATIONS THERE ARE but WHAT IS THE PROBABILITY TO HAVE THIS EVENT ?
So what ??

"exactly 3 zeros" turns out to indicate that the rest 4 digit will not include either 0 or 1.

4 odd digit for 3 slot and 4 even digit for the last slot => 4^4, and there are 4C3 ways of choosing 3 out of 4 slots.
.....
I will leave the rest to you.

Let go about it using the slots method.
i. We start by deconstructing the question into it's constituent conditions.
ii. We start with the most restrictive condition first.

So here's our question broken down into it's individual informative components:
~> A seven-digit combination lock on a safe
~> has zero exactly three times,
~> does not have the digit 1 at all.
~> What is the probability that exactly 3 of its digits are odd?

Now let's rearrange those in the order of most restrictive condition down to the least restrictive one:
1. the code has 7 digits
2. has zero exactly three times,
3. does not have the digit 1 at all.
4. exactly 3 of its digits are odd

Not much of a change! Good!
Now let's construct the solution in the same exact order :

1. the code has 7 digits

__ __ __ __ __ __ __

2. has zero exactly three times

0 0 0 __ __ __ __ (the Zeroes can be anywhere, we are only interested in the count, at this stage)

3. does not have the digit 1 at all.

Let's work with only the 4 blank slots from now on.
__ __ __ __ <-- Each of those slots can be filled with any digit among 2, 3, 4, 5, 6, 7, 8, 9 (and at this stage, the probability of choosing any digit for each individual slot is 1/8).

4. exactly 3 of its digits are odd

__ __ __ __ <-- 3 of those slots can be filled with any ODD digit among 2, 3, 4, 5, 6, 7, 8, 9 (Probability = 4/8*4/8*4/8) and the last remaining slot MUST be filled with an EVEN digit among 2, 3, 4, 5, 6, 7, 8, 9 (Probability = 4/8).

Therefore, the code must look something like:

Event 1 : (odd odd odd even) --> Probability = 4/8*4/8*4/8*4/8 = 1/16
or,
Event 2 : (odd odd even odd) --> Probability = 4/8*4/8*4/8*4/8 = 1/16
or,
Event 3 : (odd even odd odd) --> Probability = 4/8*4/8*4/8*4/8 = 1/16
or
Event 4 : (even odd odd odd) --> Probability = 4/8*4/8*4/8*4/8 = 1/16

Therefore total probability = Probability of (Event 1 + Event 2 + Event 3 + Event 4) = 1/16+1/16+1/16+1/16 = 4/16 = 1/4 -> Option D

*********
Using Formula
*********
If the probability of a certain individual event is p, then the probability of it occurring k times in n trials/iterations is:

nCk * p^k * (1 - p)^(n - k)

i.e., 4C3 * (1/2)^3 * (1 - 1/2)^1

= 4*1/16

= 1/4 -> Option D

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