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

Question

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

PKN · Answer

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)

CounterSniper · Answer

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

mj12345 · Answer

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}

Archit3110 · Answer

P = 1/6 
P=5/6
for atleast in two die ; 5/6 * 5/6 ; 25/36
1-25/36; 11/36
IMO C

Abhishek009 · Answer

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)

ScottTargetTestPrep · Answer

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

Answer: C

bumpbot · Answer

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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

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