A die is rolled 8 times. Find the probability. P(getting at least 7 even numbers)

Question

please help!!!!!!!!!!!!

Answer 1

Total number of permutations when a dice is rolled 8 times is 6 * 6 * 6 * 6 * 6 * 6 * 6 * 6 = 6⁸.

Total number of permutations for getting atleast 7 even numbers is 3 * 3 * 3 * 3 * 3 * 3 * 3 * 6

= 3⁷ * 6 where 3⁷ = choices for getting 7 times an even number and 6 = choices for getting any number.

P(getting at least 7 even numbers) = (3⁷ * 6) / 6⁸

= 3⁷ / 6⁷

= 3⁷ / (3⁷ * 2⁷)

= 1 / 2⁷

= 1 / 256.

Therefore P(getting at least 7 even numbers) = 1 / 256.

 

Answer 2

In a die there are 3 even numbers and 3 odd numbers.

Probability of getting an even number when a die is rolled = 3/6 = 1/2.

If a die is rolled 8 times, the probability of getting atleast 7 even numbers i.e., 7 times even and one time even or odd.

= (1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2) * (3/6 + 3/6)

= (1/2)⁷ * (1/2 + 1/2)

= (1/2)⁷ * (1+1)/2

= (1/2)⁷ * 2/2

= (1/2)⁷ * 1

= (1/2)⁷

= 1 / 2⁷

= 1/ 256

Therefore P(getting at least 7 even numbers) = 1 / 256.