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

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A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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12 Jun 2016, 23:43
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56% (01:21) correct 44% (02:13) wrong based on 75 sessions

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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?

A. 9/16
B. 1/2
C. 1/3
D. 1/4
E. 1/6
[Reveal] Spoiler: OA

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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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13 Jun 2016, 00:03
Number of odd digits = 4 (Since the combination excludes 1 )
Number of even digits = 4 (Since 0 occurs exactly 3 times )

Probability that exactly 3 of its digits are odd = 4c3 * (1/2)^3 * (1/2)
= 4!/3! * (1/2)^4
= 4/16
= 1/4
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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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13 Jun 2016, 02:19
Skywalker18 wrote:
Number of odd digits = 4 (Since the combination excludes 1 )
Number of even digits = 4 (Since 0 occurs exactly 3 times )

Probability that exactly 3 of its digits are odd = 4c3 * (1/2)^3 * (1/2)
= 4!/3! * (1/2)^4
= 4/16
= 1/4

Skywalker18 plz explain the formula used.. thanks

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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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13 Jun 2016, 02:36
Expert's post
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mohshu wrote:
Skywalker18 wrote:
Number of odd digits = 4 (Since the combination excludes 1 )
Number of even digits = 4 (Since 0 occurs exactly 3 times )

Probability that exactly 3 of its digits are odd = 4c3 * (1/2)^3 * (1/2)
= 4!/3! * (1/2)^4
= 4/16
= 1/4

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

Other Resources on Probability

Theory on probability problems

Data Sufficiency Questions on Probability
Problem Solving Questions on Probability

Tough Probability Questions

Probability and Geometry

Hope it helps.
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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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13 Jun 2016, 06:45
Bunuel wrote:
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?

A. 9/16
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Remaining number of digits after three '0' digits= 4

Constraint- no more '0' and no more '1'

Well, these four digits can be odd or even

1/2*1/2*1/2*1/2= 1/16

Ways to choose 3 odd digits out of total 4 digits= 4C3= 4

4* 1/16= 1/4

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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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13 Jun 2016, 07:02
Bunuel wrote:
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?

A. 9/16
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Remaining number of digits after three '0' digits= 4

Constraint- no more '0' and no more '1'

Well, these four digits can be odd or even

1/2*1/2*1/2*1/2= 1/16

Ways to choose 3 odd digits out of total 4 digits= 4C3= 4

4* 1/16= 1/4

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

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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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13 Jun 2016, 07:14
mohshu wrote:
Bunuel wrote:
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?

A. 9/16
B. 1/2
C. 1/3
D. 1/4
E. 1/6

Remaining number of digits after three '0' digits= 4

Constraint- no more '0' and no more '1'

Well, these four digits can be odd or even

1/2*1/2*1/2*1/2= 1/16

Ways to choose 3 odd digits out of total 4 digits= 4C3= 4

4* 1/16= 1/4

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
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A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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15 Jun 2016, 09:50
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 ??

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Re: A seven-digit combination lock on a safe has zero exactly three times, [#permalink]

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06 Sep 2017, 12:21
"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.

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Re: A seven-digit combination lock on a safe has zero exactly three times,   [#permalink] 06 Sep 2017, 12:21
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