A single die is rolled twice. find the probability of rolling an even number the first time and a number greater than 4 the second time.

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14-07-2018
Mathematics

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A single die is rolled twice. find the probability of rolling an even number the first time and a number greater than 4 the second time.

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midknight9 midknight9 · Answer 1 · 2018-07-14T21:53:36+02:00

So the key word we have here is 'and'. When 'and' is used, it means that both events must occur. We know that we are then going to have to multiply the individual probabilities of each event together to get the total probability of the combination.

So we first must roll and even number. A dice has six sides, so the even numbers that are possible are 2, 4, and 6. This means that out of six possible numbers to roll, we can roll 3. So our probability is:

[tex] \frac{3}{6} [/tex] or [tex] \frac{1}{2} [/tex]

And then after we roll one of those 3 even numbers, we must roll a number greater than 4. The possible numbers we can roll to get this correct are 5 and 6. This is 2 numbers out a total possible 6. So our probability is:

[tex] \frac{2}{6} [/tex] or [tex] \frac{1}{3} [/tex]

Now we need to multiply the two individual probabilities together to get the total probability:

[tex] \frac{1}{2}* \frac{1}{3}=\frac{1}{6} [/tex]

So now we know that we have a [tex] \frac{1}{6} [/tex] chance of both of these events happening consecutively.

MrRoyal MrRoyal · Answer 2 · 2021-12-03T10:47:41+01:00

Probabilities are used to determine the chances of an event

The probability of rolling an odd number the first time and a number greater than 4 the second time is 1/6

The sample space of a single die is

[tex]\mathbf{S = \{1,2,3,4,5,6\}}[/tex]

So, the total sample is 6

The odd numbers are

[tex]\mathbf{Odd = \{1,3,5\}}[/tex] --- 3 odd numbers.

So, the probability of that an odd number is rolled:

[tex]\mathbf{P(Odd) = \frac 36}[/tex]

Simplify

[tex]\mathbf{P(Odd) = \frac 12}[/tex]

The numbers greater than 4 are

[tex]\mathbf{Greater = \{5,6\}}[/tex] --- 2 numbers greater than 4.

So, the probability of that a number greater than 4 is rolled:

[tex]\mathbf{P(Greater) = \frac 26}[/tex]

Simplify

[tex]\mathbf{P(Greater) = \frac 13}[/tex]

The probability of rolling an odd number the first time and a number greater than 4 the second time is calculated as follows:

[tex]\mathbf{P = P(Odd) \times P(Greater)}[/tex]

So, we have:

[tex]\mathbf{P = \frac 12 \times \frac 13}[/tex]

[tex]\mathbf{P = \frac 16}[/tex]

Hence, the probability is 1/6