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Events & Promotions

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After the LuckyAir flight arrived in Las Vegas, John went to

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Senior Manager
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After the LuckyAir flight arrived in Las Vegas, John went to [#permalink]

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New post 19 Oct 2005, 13:09
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After the LuckyAir flight arrived in Las Vegas, John went to
the strip to gamble. At the craps table, John bet $20 on
boxcars (two sixes). If John rolls two fair, six-sided dice,
what is the probability that he will roll two sixes?

A. 2/12
B. 1/8
C. 1/2
D. 1/24
E. 1/36
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New post 19 Oct 2005, 14:46
Probability of rolling a 6 on the 1st dice = 1/6
Probability of rolling a 6 on the 2nd dice = 1/6

Since these are not OR probabilities, meaning they both have to occur, you mulitply them.

1/6 * 1/6 = 1/36
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New post 19 Oct 2005, 18:53
OA is E.

Here is the detailed explanation:
If two events are independent, the probability of both events occurring is the product of the two individual probabilities. The rolling of each die is an independent event. The outcome of one die does not affect the outcome of the other.

The probability of an event occurring is the number of possible outcomes for the event divided by the total number of possible outcomes. Their is only one outcome out of six that will produce a six when you roll a die. Thus, the probability of one die rolling a six is 1/6.

The probability of rolling two sixes is (1/6)*(1/6) or 1/36.
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New post 19 Oct 2005, 19:04
P(two six) = P(six on die 1) * P(six on die 2) = 1/36
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Re: PS: Probability - Dice + Vegas [#permalink]

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New post 19 Oct 2005, 21:56
sudhagar wrote:
After the LuckyAir flight arrived in Las Vegas, John went to
the strip to gamble. At the craps table, John bet $20 on
boxcars (two sixes). If John rolls two fair, six-sided dice,
what is the probability that he will roll two sixes?

A. 2/12
B. 1/8
C. 1/2
D. 1/24
E. 1/36


These are independent events. So it is the product of each of the probablities.

Hence it is (1/6) * (1/6) = 1/36
Re: PS: Probability - Dice + Vegas   [#permalink] 19 Oct 2005, 21:56
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