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# If the probability is 1/6 that a die will turn up as one of the number

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If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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05 Aug 2018, 10:47
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35% (medium)

Question Stats:

64% (01:11) correct 36% (00:54) wrong based on 75 sessions

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If the probability is 1/6 that a die will turn up as one of the numbers from 1 to 6, what is the probability that in two rolls of that die, at least one of the rolls turns up as 6?
A) 1/36
B) 1/6
C) 11/36
D) 25/36
E) 5/6

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Re: If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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05 Aug 2018, 11:11
gmatbusters wrote:
If the probability is 1/6 that a die will turn up as one of the numbers from 1 to 6, what is the probability that in two rolls of that die, at least one of the rolls turns up as 6?
A) 1/36
B) 1/6
C) 11/36
D) 25/36
E) 5/6

Let P(A) denotes probability that roll turns up as 6.

Probability that at least one of the rolls turn up as 6= probability that two of the rolls turn up as 6+Probability of that 1st die turns up 6 and 2nd die turn a non-6+ probability that 1st die turns up as non-6 and 2nd die turns up as 6
=P(AA)+P(AA')+P(A'A)
=1/6*1/6+(1/6)(1-1/6)+(1-1/6)*(1/6)
=1/36+2*(1/6)*(5/6)
=1/36+10/36
=11/36
Ans. (C)
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Re: If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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07 Aug 2018, 13:01
gmatbusters wrote:
If the probability is 1/6 that a die will turn up as one of the numbers from 1 to 6, what is the probability that in two rolls of that die, at least one of the rolls turns up as 6?
A) 1/36
B) 1/6
C) 11/36
D) 25/36
E) 5/6

1. The first roll is a 6 ,but the second is not
+
2. The first roll is not a 6 , but the second roll is
+
3. Both are 6

we therefore have
(1/6 * 5/6) + ( 5/6 * 1/6 ) + (1/6 * 1/6)=11/36
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Re: If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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07 Aug 2018, 21:25
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Probability that we'll roll a 6 = $$\frac{1}{6}$$

Probability that we'll roll anything but 6 =$$\frac{5}{6}$$

For two throws, probability that we won't roll a 6 in either throw = $$\frac{5}{6}*\frac{5}{6}$$ = $$\frac{25}{36}$$

Probability that we roll 6 in at least one throw = $$1 - \frac{25}{36}$$ = $$\frac{11}{36}$$
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Re: If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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22 May 2019, 08:34
gmatbusters wrote:
If the probability is 1/6 that a die will turn up as one of the numbers from 1 to 6, what is the probability that in two rolls of that die, at least one of the rolls turns up as 6?
A) 1/36
B) 1/6
C) 11/36
D) 25/36
E) 5/6

P = 1/6
P=5/6
for atleast in two die ; 5/6 * 5/6 ; 25/36
1-25/36; 11/36
IMO C
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Re: If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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22 May 2019, 12:16
gmatbusters wrote:
If the probability is 1/6 that a die will turn up as one of the numbers from 1 to 6, what is the probability that in two rolls of that die,at least one of the rolls turns up as 6?
A) 1/36
B) 1/6
C) 11/36
D) 25/36
E) 5/6

The highlighted part includes -

1. First dice is 6
2. Second dice is 6
3. Both dices are 6

So, The required probability is 1 - Prob that both the dice doesn't result in 6

ie, $$1 - \frac{5}{6}*\frac{5}{6}$$

Or, Required probability is $$1 - \frac{25}{36} = \frac{11}{36}$$, Answer must be (C)
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Re: If the probability is 1/6 that a die will turn up as one of the number  [#permalink]

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27 May 2019, 05:21
gmatbusters wrote:
If the probability is 1/6 that a die will turn up as one of the numbers from 1 to 6, what is the probability that in two rolls of that die, at least one of the rolls turns up as 6?
A) 1/36
B) 1/6
C) 11/36
D) 25/36
E) 5/6

P(at least one 6) = 1 - P(no sixes)

P(at least one 6) = 1 - 5/6 x 5/6 = 1 - 25/36 = 11/36

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Re: If the probability is 1/6 that a die will turn up as one of the number   [#permalink] 27 May 2019, 05:21
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