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Answer :
The value of x is 5042 .
To find the probability of the man hitting the target at least twice out of 7 shots, we can use the binomial probability formula:
[tex]\[ P(X \geq 2) = 1 - P(X < 2) \][/tex]
Where:
[tex]\( P(X \geq 2) \)[/tex] is the probability of hitting the target at least twice.
[tex]\( P(X < 2) \)[/tex] is the probability of hitting the target less than twice.
Let's calculate P(X < 2) first.
[tex]\[ P(X < 2) = P(X = 0) + P(X = 1) \][/tex]
Where:
P(X = 0) is the probability of not hitting the target at all.
P(X = 1) is the probability of hitting the target once.
The formula for the probability of hitting the target k times out of n shots is given by the binomial distribution formula:
[tex]\[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \][/tex]
Where:
n is the number of trials (in this case, shots),
k is the number of successes (in this case, hitting the target),
p is the probability of success on each trial (in this case, hitting the target),
[tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, also known as "n choose k".
Given that p = 0.25 , n = 7 :
[tex]\[ P(X = 0) = \binom{7}{0} \times 0.25^0 \times (1-0.25)^{7-0} \]\[ P(X = 0) = 1 \times 1 \times 0.75^7 = 0.1335 \]\[ P(X = 1) = \binom{7}{1} \times 0.25^1 \times (1-0.25)^{7-1} \]\[ P(X = 1) = 7 \times 0.25 \times 0.75^6 = 0.2503 \]\[ P(X < 2) = 0.1335 + 0.2503 = 0.3838 \][/tex]
Now, we can find [tex]\( P(X \geq 2) \):[/tex]
[tex]\[ P(X \geq 2) = 1 - P(X < 2) = 1 - 0.3838 = 0.6162 \][/tex]
To express[tex]\( P(X \geq 2) \)[/tex] as a fraction with a denominator of 8192, we can multiply both the numerator and the denominator by 8192 :
[tex]\[ P(X \geq 2) = \frac{0.6162 \times 8192}{1 \times 8192} \]\[ P(X \geq 2) = \frac{5041.7664}{8192} \][/tex]
So, the value of x is 5042 .
Question
The probability of a man hitting a target is 0.25. He shoots 7 times. the probability of his hitting at least twice is x/ 8192 , then the value of x is ___
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