The space shuttle is launched 9 times a year with a break down occurring 1/10 of the time. What is the probability that two randomly selected space shuttles launched 9 times a year and both having accidents occurring in that year?

Answer 1

Let's first find the probability of a single space shuttle having an accident in a year.

There are 9 launches in a year, and the probability of a breakdown in each launch is 1/10. So the probability of at least one breakdown occurring in the year is 1 - (the probability of no breakdowns occurring in the year).

The probability of no breakdowns occurring in a single launch is 1 - 1/10 = 9/10. The probability of no breakdowns occurring in all 9 launches is (9/10)^9.

So, the probability of at least one breakdown occurring in a year for a single space shuttle is 1 - (9/10)^9.

Now, we need to find the probability that both randomly selected space shuttles have accidents in the same year.

The probability of both events happening is the product of their individual probabilities: (1 - (9/10)^9) * (1 - (9/10)^9) = (1 - (9/10)^9)^2.

Calculating this value, we get (1 - (9/10)^9)^2 ≈ 0.6108 or 61.08%. So, there is approximately a 61.08% probability that two randomly selected space shuttles launched 9 times a year will both have accidents occurring in that year.