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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?

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

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13 Jun 2016, 01:03

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 Answer D
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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 \(\frac{5!}{3!2!}\).

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

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13 Jun 2016, 07:45

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

D is the answer
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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, 08:02

Divyadisha 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

D is the answer

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

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

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13 Jun 2016, 08:14

mohshu wrote:

Divyadisha 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

D is the answer

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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I welcome critical analysis of my post!! That will help me reach 700+

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

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15 Jun 2016, 10:50

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 ??

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

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06 Sep 2017, 13:21

"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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