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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 probability that the sum of the dice is 5 given that the outcome is not 6 is (Type an integer or a simplified fraction.)

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LammettHash

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

The required probability that the sum of the dice is 5 given that the outcome is not 6 is 4/31

What is probability?

  • Probability is the chance of happening of an event.
  • Probability is always ≤ 1

What is conditional probability?

Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome.

How to find the probability that the sum of the dice is 5 given that the outcome is not 6 ?

According to the problem,

  • A pair of dice is rolled.

The problem can be mathematically written as,

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

Now,

  • When a pair of dice is rolled, there are 36 possible outcomes.
  • So the sample space is 36.

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.

Again,

  • 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)
  • So there are 31 ways of not getting a sum of 6,

So, P(sum is not 6) = 31/36.

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

This is the required probability.

Find more about "Probability" here : https://brainly.com/question/25870256

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