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?
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.
Well, let's break it down. The probability of a breakdown occurring during a launch is 1/10, which means the probability of a successful launch is 9/10.
Since we are looking for the probability of two shuttles having accidents in the same year, we need to multiply these probabilities together.
So, the probability of the first shuttle having an accident is 1/10, and the probability of the second shuttle having an accident is also 1/10.
Multiplying these probabilities together: (1/10) * (1/10) = 1/100.
Therefore, the probability that both randomly selected shuttles from the 9 launches in a year have accidents is 1/100.
However, I must clarify that these calculations are based purely on the given information. Real-life probabilities might differ.
To determine the probability of two randomly selected space shuttles both having accidents in a given year, we need to find the product of the probabilities of each individual accident occurring.
Given that a breakdown occurs 1/10 of the time, the probability of a single shuttle having an accident is 1/10. Since there are two shuttles being considered, we need to multiply this probability by itself for each shuttle.
Therefore, the probability that both shuttles have accidents in a given year is (1/10) * (1/10), which simplifies to 1/100.
Hence, the probability that two randomly selected space shuttles launched 9 times a year both have accidents occurring in that year is 1/100.
To find the probability that two randomly selected space shuttles launched 9 times a year both have accidents occurring in that year, we need to consider the probability of one space shuttle having an accident and multiply it by the probability of the other space shuttle having an accident.
First, let's find the probability of a single space shuttle having an accident. We are given that the breakdown occurs 1/10 (or 0.1) of the time. So, the probability of a space shuttle having an accident is 0.1.
Since we have two space shuttles and we want both of them to have accidents, we multiply the probabilities together:
0.1 * 0.1 = 0.01
Therefore, the probability that two randomly selected space shuttles launched 9 times a year both have accidents occurring in that year is 0.01 or 1%.