1. What is the probability of exactly 3 light bulbs burning out within a year, given that there are 8 light bulbs
1. What is the probability of exactly 3 light bulbs burning out within a year, given that there are 8 light bulbs in the store and each has a 0.1 probability of burning out within a year?
2. Three archers are shooting at a target. The probabilities of hitting the target for each archer are 0.76, 0.72, and 0.8, respectively. After they all shoot simultaneously, two arrows are found on the target. What is the probability that the second archer hit the target?
3. Three archers are shooting at a target. The probabilities of hitting the target for each archer are 0.76, 0.72, and 0.8, respectively. After they all shoot simultaneously, two arrows are found on the target. What is the probability
2. Three archers are shooting at a target. The probabilities of hitting the target for each archer are 0.76, 0.72, and 0.8, respectively. After they all shoot simultaneously, two arrows are found on the target. What is the probability that the second archer hit the target?
3. Three archers are shooting at a target. The probabilities of hitting the target for each archer are 0.76, 0.72, and 0.8, respectively. After they all shoot simultaneously, two arrows are found on the target. What is the probability
Веселый_Смех 38
Sure! Let"s solve these problems step by step.1. To find the probability of exactly three light bulbs burning out within a year, given that there are 8 light bulbs in the store and each has a 0.1 probability of burning out within a year, we can use the binomial distribution formula.
The probability of an event occurring (p) is 0.1, and we want exactly 3 successes (k) out of 8 trials (n).
The formula for the probability mass function (PMF) of the binomial distribution is:
\[P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\]
where \(\binom{n}{k}\) represents the binomial coefficient, which is the number of ways to choose k items out of n items.
The binomial coefficient can be calculated as:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
Let"s substitute the values into the formula:
\[P(X=3) = \binom{8}{3} \cdot (0.1)^3 \cdot (1-0.1)^{8-3}\]
Simplifying further:
\[P(X=3) = \frac{8!}{3!(8-3)!} \cdot (0.1)^3 \cdot (0.9)^5\]
Evaluate the binomial coefficient:
\[\binom{8}{3} = \frac{8!}{3! \cdot 5!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56\]
Substituting the values:
\[P(X=3) = 56 \cdot (0.1)^3 \cdot (0.9)^5\]
Calculating this expression we get the final answer:
\[P(X=3) = 0.057395628\]
Therefore, the probability of exactly 3 light bulbs burning out within a year is approximately 0.057.
2. To find the probability that the second archer hit the target, given that two arrows are found on the target, we can use conditional probability.
Let"s define the events A as the second archer hitting the target and B as two arrows being found on the target.
We need to find \(P(A|B)\), which represents the probability of A happening given that B has occurred.
Using Bayes" theorem, we have:
\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\]
We know that \(P(B|A)\) is the probability of two arrows being found on the target given that the second archer hit the target. This is equal to 1 since if the second archer hit the target, we will always find two arrows on the target.
We also know that \(P(A)\) is the probability of the second archer hitting the target, which is given as 0.72.
\(P(B)\) is the probability of two arrows being found on the target, regardless of who hit it. This can be calculated using the law of total probability.
\[P(B) = P(B|A) \cdot P(A) + P(B|A") \cdot P(A")\]
where A" represents the event of the second archer not hitting the target.
\(P(B|A")\) is the probability of two arrows being found on the target given that the second archer did not hit the target. This is equal to the product of the probabilities of the first and third archers hitting the target (0.76 and 0.8, respectively).
Substituting the given values:
\[P(B) = 1 \cdot 0.72 + (0.76) \cdot (0.8) \cdot (1-0.72)\]
Simplifying:
\[P(B) = 0.72 + 0.2176\]
\[P(B) = 0.9376\]
Finally, substituting all the values into Bayes" theorem:
\[P(A|B) = \frac{1 \cdot 0.72}{0.9376}\]
Calculating this expression we get the final answer:
\[P(A|B) = 0.76821192\]
Therefore, the probability that the second archer hit the target, given that two arrows are found on the target, is approximately 0.768.
3. Similarly to the previous problem, to find the probability that the second archer hit the target, we can use conditional probability.
Let"s define the events A as the second archer hitting the target and B as two arrows being found on the target.
We need to find \(P(A|B)\), which represents the probability of A happening given that B has occurred.
Using Bayes" theorem, we have:
\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\]
We know that \(P(B|A)\) is the probability of two arrows being found on the target given that the second archer hit the target. This is equal to 1 since if the second archer hit the target, we will always find two arrows on the target.
We also know that \(P(A)\) is the probability of the second archer hitting the target, which is given as 0.72.
\(P(B)\) is the probability of two arrows being found on the target, regardless of who hit it. This can be calculated using the law of total probability.
\[P(B) = P(B|A) \cdot P(A) + P(B|A") \cdot P(A")\]
where A" represents the event of the second archer not hitting the target.
\(P(B|A")\) is the probability of two arrows being found on the target given that the second archer did not hit the target. This is equal to the product of the probabilities of the first and third archers hitting the target (0.76).
Substituting the given values:
\[P(B) = 1 \cdot 0.72 + (0.76) \cdot (1-0.72)\]
Simplifying:
\[P(B) = 0.72 + 0.2128\]
\[P(B) = 0.9328\]
Finally, substituting all the values into Bayes" theorem:
\[P(A|B) = \frac{1 \cdot 0.72}{0.9328}\]
Calculating this expression we get the final answer:
\[P(A|B) = 0.77241379\]
Therefore, the probability that the second archer hit the target, given that two arrows are found on the target, is approximately 0.772.