The probability that more than 1 vessel transporting nuclear weapons was destroyed is approximately 0.2848, rounded to four decimal places.
To solve this problem, we can use the complement probability.
The probability of more than 1 vessel carrying nuclear weapons being destroyed is equal to 1 minus the probability of 0 or 1 vessel carrying nuclear weapons being destroyed.
Let's calculate the probability of 0 or 1 vessel carrying nuclear weapons being destroyed:
The probability of a vessel carrying nuclear weapons being destroyed in a single attempt is the ratio of the number of vessels carrying nuclear weapons to the total number of vessels.
In this case, it is 9/23.
Therefore, the probability of a vessel carrying nuclear weapons not being destroyed in a single attempt is 1 - 9/23 = 14/23.
Now, we can calculate the probability of 0 or 1 vessel carrying nuclear weapons being destroyed in 10 attempts.
We can use the binomial probability formula:
P(X≤1)= (
) (
)^10 + (
) (
)^9 (
)
Using a calculator or statistical software, we find this probability to be approximately 0.7152.
Finally, the probability of more than 1 vessel carrying nuclear weapons being destroyed is:
P(More than 1)=1−P(X≤1)
P(More than 1)≈1−0.7152≈0.2848
So, the probability that more than 1 vessel transporting nuclear weapons was destroyed is approximately 0.2848, rounded to four decimal places.