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____ [38]
3 years ago
10

When a pair of dice is rolled, what is the probability that the sum of the dice is 5, given that the outcome is not 6? The proba

bility that the sum of the dice is 5 given that the outcome is not 6 is (Type an integer or a simplified fraction.)
Mathematics
1 answer:
Andrew [12]3 years ago
4 0

By definition of conditional probability,

P(sum is 5 | sum is not 6) = P(sum is 5 AND sum is not 6) / P(sum is not 6)

There are 4 ways of rolling the dice to get a sum of 5 out of 36 possible outcomes:

(1, 4), (2, 3), (3, 2), (4, 1)

so P(sum is 5) = 4/36 = 1/9.

There are 5 ways of getting a sum of 6 out of 36 possible outcomes:

(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)

hence 31 ways of *not* getting a sum of 6, so P(sum is not 6) = 31/36.

The sum is 5 and *not* 6 simultaneously for the 4 ways already listed above, so the intersection of these two events is those 4 ways, which means

P(sum is 5 AND sum is not 6) = 1/9

Then

P(sum is 5 | sum is not 6) = (1/9) / (31/36) = 4/31

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erastovalidia [21]

Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

This means that n = 16

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that p = 0.5

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0

P(X = 1) = C_{16,1}.(0.5)^{1}.(0.5)^{15} = 0.0002

P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018

P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085

Then

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0002 + 0.0018 + 0.0085 = 0.0105

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

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Answer:

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Step-by-step explanation:

Do the following:

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Plugging this into a calculator, you will get:

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Answer:

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Step-by-step explanation:

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