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If a six sided die is rolled three times, what is the probability of

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If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 24 Jul 2019, 02:57
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If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post Updated on: 24 Jul 2019, 16:55
# of favourable cases = 3*3*3*6
# of total cases = 6*6*6
Probability = (3*3*3*6)/(6*6*6) = 3/4

IMO D

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Originally posted by Kinshook on 24 Jul 2019, 03:01.
Last edited by Kinshook on 24 Jul 2019, 16:55, edited 1 time in total.
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If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post Updated on: 24 Jul 2019, 04:23
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The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8
Also, number of ways of rolling a six sided die like: (EvenOddOdd, Odd,Even,Odd, OddOddEven)—3 ways —> 3*1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8
Also, number of ways of rolling a six sided die like: (EvenOddEven, Even,Even,Odd, OddEvenEven)—3 ways—> 3*1/8

3/8+3/8= 3/4
The answer choice is D

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Originally posted by lacktutor on 24 Jul 2019, 03:05.
Last edited by lacktutor on 24 Jul 2019, 04:23, edited 1 time in total.
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 24 Jul 2019, 03:42
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Probability of getting at least one EVEN and at least one ODD \(=\) Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability of getting 3 EVEN's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability that the three rolls are all ODD or all EVEN \(= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}\)

Probability that the three rolls are NOT all ODD or NOT all EVEN \(= 1-\frac{1}{4} = \frac{3}{4}\)

Answer is (D)
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 24 Jul 2019, 03:53
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lacktutor wrote:
The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8

1/8+1/8= 1/4
The answer choice is B.

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Dear lacktutor

What you appear to have failed to take into account is - the number of ways you can choose the one die that would be either even or odd, i.e. 3C1

Taking that into account, your answer should be as follows:

3C1*3/6*3/6*3/6 (for 1 odd and 2 even) and 3C2*3/6*3/6*3/6 (for 2 odd and 1 even)

3*1/8 + 3*1/8 = 6/8 or 3/4, and hence, (D) is the correct answer choice

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 24 Jul 2019, 04:09
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Total outcomes = 6*6*6= 216
If all the numbers are odd 3*3*3 = 27
If all the numbers are even 3*3*3= 27
So sum of all even and all odd outcomes = 54
Probability of all odd or all even 54/216 = 1/4
Hence probability of at least one even and one add = 3/4
Hence D

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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 24 Jul 2019, 04:27
RJ7X0DefiningMyX wrote:
lacktutor wrote:
The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8

1/8+1/8= 1/4
The answer choice is B.

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Dear lacktutor

What you appear to have failed to take into account is - the number of ways you can choose the one die that would be either even or odd, i.e. 3C1

Taking that into account, your answer should be as follows:

3C1*3/6*3/6*3/6 (for 1 odd and 2 even) and 3C2*3/6*3/6*3/6 (for 2 odd and 1 even)

3*1/8 + 3*1/8 = 6/8 or 3/4, and hence, (D) is the correct answer choice

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION


Yes, you’re right. I didn’t add the number of combinations within Even and Odd.
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 24 Jul 2019, 06:29
Bunuel wrote:
If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number?

A. 1/8
B. 1/4
C. 1/2
D. 3/4
E. 7/8


we are considering cases w/o replacement
so atleast even ; 3/6*3/6*3/6 ; 1/8
and odd ; 1/8
total cases ; 1/8+1/8 ; 1/4
1-1/4 ; 3/4
IMO D
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If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 05 Jan 2020, 05:10
lacktutor wrote:
The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8
Also, number of ways of rolling a six sided die like: (EvenOddOdd, Odd,Even,Odd, OddOddEven)—3 ways —> 3*1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8
Also, number of ways of rolling a six sided die like: (EvenOddEven, Even,Even,Odd, OddEvenEven)—3 ways—> 3*1/8

3/8+3/8= 3/4
The answer choice is D

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Why we are adding the probabilities at last to get the final resultant 3/4 when the question itself ask to calculate the probability of getting atleast one even number and atleast one odd number ?

*AND* is clearly mentioned : so we should perform multiplication, isn’t it ???

Please correct me
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If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 05 Jan 2020, 05:14
firas92 wrote:
Probability of getting at least one EVEN and at least one ODD \(=\) Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability of getting 3 EVEN's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability that the three rolls are all ODD or all EVEN \(= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}\)

Probability that the three rolls are NOT all ODD or NOT all EVEN \(= 1-\frac{1}{4} = \frac{3}{4}\)

Answer is (D)


Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Please correct me Bunuel VeritasKarishma

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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 05 Jan 2020, 22:23
LeenaSai wrote:
firas92 wrote:
Probability of getting at least one EVEN and at least one ODD \(=\) Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability of getting 3 EVEN's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability that the three rolls are all ODD or all EVEN \(= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}\)

Probability that the three rolls are NOT all ODD or NOT all EVEN \(= 1-\frac{1}{4} = \frac{3}{4}\)

Answer is (D)


Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Please correct me Bunuel VeritasKarishma

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The probability of two independent events occurring together is given by the product of their probabilities. These two events (at least one odd and at least one even) are not independent. So this method is not correct.
From the total, if you remove the probability of all odd and all even (mutually exclusive events so you just subtract their probabilities one after the other) event, you get that the remaining cases will have at least one odd and one even outcome.
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 05 Jan 2020, 22:38
VeritasKarishma wrote:
LeenaSai wrote:
firas92 wrote:
Probability of getting at least one EVEN and at least one ODD \(=\) Probability that the three rolls are NOT all ODD or NOT all EVEN

Probability of getting 3 ODD's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability of getting 3 EVEN's \(= (\frac{3}{6})*(\frac{3}{6})*(\frac{3}{6}) = \frac{1}{8}\)

Probability that the three rolls are all ODD or all EVEN \(= \frac{1}{8}+\frac{1}{8} = \frac{1}{4}\)

Probability that the three rolls are NOT all ODD or NOT all EVEN \(= 1-\frac{1}{4} = \frac{3}{4}\)

Answer is (D)


Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Please correct me Bunuel VeritasKarishma

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The probability of two independent events occurring together is given by the product of their probabilities. These two events (at least one odd and at least one even) are not independent. So this method is not correct.
From the total, if you remove the probability of all odd and all even (mutually exclusive events so you just subtract their probabilities one after the other) event, you get that the remaining cases will have at least one odd and one even outcome.



Now it’s pretty clear VeritasKarishma ?
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Re: If a six sided die is rolled three times, what is the probability of  [#permalink]

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New post 05 Jan 2020, 22:56
E O E/O . This anagram wheter EOO or EOE cn be permuted in 3!/2! ways = 3 ways.
Further,

getting probability of (E) (O) (E/O) = 1/2 * 1/2 * 6/6 = 1/4 and now we will permute this EOE or EEO in 3 ways so
multiply it with obtained probability = 1/4 * 3 = 3/4 ----> D
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Re: If a six sided die is rolled three times, what is the probability of   [#permalink] 05 Jan 2020, 22:56
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