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A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di

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Bunuel
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A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 03:12
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A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

A. 1/6
B. 625/1296
C. 649/1296
D. 671/1296
E. 2/3

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D
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 03:23
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Bunuel wrote:
A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

A. 1/6
B. 625/1296
C. 649/1296
D. 671/1296
E. 2/3


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Method 1:

Atleast one of the die shows 6 i.e.

- Exactly 1 die of 4 shows 6 - Outcomes = 4C1*5^3 = 4*125 = 500
- Exactly 2 dice of 4 show 6 - Outcomes = 4C2*5^2 = 6*25 = 150
- Exactly 3 dice of 4 show 6 - Outcomes = 4C3*5 = 4*5 = 20
- Exactly 4 dice of 4 show 6 - Outcomes = 4C4 = 1

Total Outcomes = \(6^4\)

Probability \(= \frac{(500+150+20+1)}{6^4} = \frac{671}{1296}\)

Answer: Option D


Method 2:

Probability of atleast 1 six \(= 1- (Probability of none Six) = 1-(\frac{5^4}{6^4}) = 1- (\frac{625}{1296}) = \frac{671}{1296}\)

Answer: Option D
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 03:16
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Probability of atleast one 4 = 1 - Probability of zero 4
Probability of atleast one 4 = 1 - (5/6)^4
Probability of atleast one 4 = 1 - 625/1296
Probability of atleast one 4 = 671/1296

Answer: D
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 03:28
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Bunuel wrote:
A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

A. 1/6
B. 625/1296
C. 649/1296
D. 671/1296
E. 2/3


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Probability of getting a 6 on any die=1/6
so P(not getting 6)=1-1/6=5/6
if rolled 4 times Probability of not getting 6 in any of the 4 dies= \(5^4/6^4\)
P(at least one 6)=1-625/1296
=671/1296
D:)
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 03:50
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Bunuel wrote:
A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

A. 1/6
B. 625/1296
C. 649/1296
D. 671/1296
E. 2/3


Solution


    • We know that, if a fair dice is rolled n times, the total number of possible outcomes \(= 6^n\)
      o So, total possible outcomes when a fair dice is rolled 4 times \(= 6^4 = 1296\)
    • Also, total number of possible outcomes in which 6 won’t show up on any roll \(= 5*5*5*5 = 625\)
      o Therefore, P(Not getting 6 on any roll) \(= \frac{625}{1296}\)
    • Hence, P(getting 6 on at least one roll) = 1 – P(Not getting 6 on any roll) \(= 1 - \frac{625}{1296} = \frac{671}{1296} \)
Thus, the correct answer is Option D.
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 03:50
Kindly see the attachment. For a minute I thought that will it be right to the concept of (1- event not occurring), but then I realized that 6 can occur at any place, so the concept can work here.

TIME: 01:25
IMO D
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 06:10
Explanation:

Probability of atleast one six =1−(Probability of zero Six)
= 1 − (5^4/6^4)
= 1− (625/1296)
= 671/1296

IMO-D
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 06:41
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Atleast one time 6 during time dice roll,
probability of at least one time means, either 1 time or 2 time or 3 time or all the 4 time.

This can be calculated by finding probability of no 6 in all 4 dice roll & subtract from 1. probability of getting a 6 is 1/6 & not getting a 6 during one time dice roll is 5/6.

1 - (5/6)*(5/6)*(5/6)*(5/6)
671/1296
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   01 Jul 2020, 07:10
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atleast -> P(6) = 1 - P(Not 6)

P(6) = 1 - [(5/6)*(5/6)*(5/6)*(5/6)] -> the die is rolled 4 times
P(6) = 1- (625/1296)
P(6) = 671/1296

Answer -> D
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   05 Jul 2020, 05:52
Bunuel wrote:
A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

A. 1/6
B. 625/1296
C. 649/1296
D. 671/1296
E. 2/3


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Could someone tell me what is wrong with the below approach :

(1/6*5/6*5/6*5/6)(4!/3!) + (1/6*1/6*5/6*5/6)(4!/(2!*2!)) + (1/6*1/6*1/6*5/6)(4!/3!) + (1/6*1/6*5/6*1/6)
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   05 Jul 2020, 18:17
Factorial part added with the 1st three are incorrect. There should not be a factorial part in that expression.
Rest is correct!

(6 is 1st dice not other 3 times) +(6 in 1st 2 times of dice not other 2 times)+ (6 in 1st 3 times of dice not 4th time) + (6 in all 4 times of dice)

Hope it helps!

Posted from my mobile device
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   06 Jul 2020, 05:57
keshavmishra wrote:
Factorial part added with the 1st three are incorrect. There should not be a factorial part in that expression.
Rest is correct!

(6 is 1st dice not other 3 times) +(6 in 1st 2 times of dice not other 2 times)+ (6 in 1st 3 times of dice not 4th time) + (6 in all 4 times of dice)

Hope it helps!

Posted from my mobile device


Tried it without including the factorial expressions, that doesn't lead to any of the answer choices provided. Could you show me your steps more elaborately?
Moreover I'd like to know why including the factorial bit is incorrect. I assumed that that is relevant because the required combo could occur in multiple ways.
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink] New post   15 Oct 2022, 13:12
1-(5/6)^4.

Answer D.
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Re: A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the di [#permalink]
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