Question

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

Answer 1

Answer: 1/6

Step-by-step explanation:

A die has 6 numbers which are 1, 2, 3, 4, 5 and 6.

Odd numbers in a die = 1, 3 and 6

Numbers greater than 4 = 5 and 6

Probability of rolling an odd number = 3/6 = 1/2

Probability of rolling a number greater than 4 = 2/6 = 1/3

We then multiply both values gotten. This will be:

= 1/2 × 1/3

= 1/6

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

Answer 2

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

$\mathbf{S = \{1,2,3,4,5,6\}}$

So, the total sample is 6

The odd numbers are

$\mathbf{Odd = \{1,3,5\}}$ --- 3 odd numbers.

So, the probability of selecting an odd number is:

$\mathbf{P(Odd) = \frac 36}$

Simplify

$\mathbf{P(Odd) = \frac 12}$

The numbers greater than 4 are

$\mathbf{Greater = \{5,6\}}$ --- 2 numbers greater than 4.

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

$\mathbf{P(Greater) = \frac 26}$

Simplify

$\mathbf{P(Greater) = \frac 13}$

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

$\mathbf{P = P(Odd) \times P(Greater)}$

So, we have:

$\mathbf{P = \frac 12 \times \frac 13}$

$\mathbf{P = \frac 16}$

Hence, the probability is 1/6

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