3 years ago

PIUWawilities

Mathematics

1 answer:

blondinia [14]3 years ago

3 0

Answer:

23.44%

Step-by-step explanation:

The probability of getting a 4 on the first 2 throws and different numbers on the last 5 throws = 1/6 * 1/6 * (5/6)^5

= 0.01116

There are 7C2 ways of the 2 4's being in different positions

= 7*6 / 2 = 21 ways.

So the required probability = 0.01116 * 21

= 0.2344 or 23.44%.

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