Question

A fair dice is thrown 3 times. The probability that the product of the three outcomes is a prime number

Answer 1

Answer:

Step-by-step explanation:

Assuming you have a standard die of six sides, with each side containing a number 1–6, then there are 3 prime number: 2, 3, 5.

The probability of rolling 3 numbers out of 6 is 3/6 or 1/2.

Repeating this 3 times is 1/2 x 1/2 x 1/2 = 1/8.

Answer 2

Answer:

The probability would be $\dfrac{1}{24}$

Step-by-step explanation:

When a die is thrown three times,

Then the total possible outcomes, n(S) = 6 × 6 × 6 = 216,

The total ways of getting a product of the three outcomes is a prime number, say E = {(1, 2, 1), (1, 3, 1), (1, 5, 1), (1, 1, 2), (1, 1, 3), (1, 1, 5), (2, 1, 1), (3, 1, 1), (5, 1, 1)}

i.e. n(E) = 9,

Hence, the probability of  getting the product of prime number,

$P(E) =\frac{n(E)}{n(S)}$

$=\frac{9}{216}$

$=\frac{1}{24}$

#Learn more:

When a dice is thrown, find the probability of not getting a prime number.

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