In a video game, the player conducts shooting training. Each shot has a probability of 0.7 to hit and a probability of 0.3 to miss. The player shoots three times in a row. Please calculate the probability of the following situations:Exactly one hit（Round to three decimal places）？
Question
In a video game, the player conducts shooting training. Each shot has a probability of 0.7 to hit and a probability of 0.3 to miss. The player shoots three times in a row. Please calculate the probability of the following situations:Exactly one hit（Round to three decimal places）？
Solution
This is a binomial probability problem. The formula for binomial probability is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the number of combinations of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case:
- n = 3 (the number of shots)
- k = 1 (the number of hits we want to get)
- p = 0.7 (the probability of hitting a shot)
Substituting these values into the formula, we get:
P(X=1) = C(3, 1) * (0.7^1) * ((1-0.7)^(3-1))
Calculating the above expression gives a probability of 0.189 when rounded to three decimal places.
So, the probability of getting exactly one hit when shooting three times is 0.189.
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