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Answer :
The probability of winning no more than 8 times out of 10 plays of the arcade game is approximately 0.6394.
Here, we have,
To calculate the probability of winning no more than 8 times out of 10 plays of the arcade game, we need to find the cumulative probability of winning 0, 1, 2, 3, 4, 5, 6, 7, and 8 times.
We'll use the binomial distribution formula to do this:
The binomial distribution formula for a probability of k successes in n trials is given by:
[tex]P(X = k) = C(n, k) * p^k * (1 - p)^{(n - k)[/tex]
where:
P(X = k) is the probability of exactly k successes,
n is the total number of trials (10 in this case),
k is the number of successes we want to calculate the probability for (0 to 8 in this case),
p is the probability of winning (0.632 in this case),
C(n, k) is the binomial coefficient, calculated as
C(n, k) = n! / (k! * (n - k)!).
Now, let's calculate the probabilities for each value of k and then sum them up to find the cumulative probability:
P(X ≤ 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
now, we have,
[tex]P(X = 0) = C(10, 0) * (0.632)^0 * (1 - 0.632)^{(10 - 0)}\\P(X = 1) = C(10, 1) * (0.632)^1 * (1 - 0.632)^{(10 - 1)}\\P(X = 2) = C(10, 2) * (0.632)^2 * (1 - 0.632)^{(10 - 2)}\\P(X = 3) = C(10, 3) * (0.632)^3 * (1 - 0.632)^{(10 - 3)}\\P(X = 4) = C(10, 4) * (0.632)^4 * (1 - 0.632)^{(10 - 4)}\\[/tex]
[tex]P(X = 5) = C(10, 5) * (0.632)^5 * (1 - 0.632)^{(10 - 5)}\\P(X = 6) = C(10, 6) * (0.632)^6 * (1 - 0.632)^{(10 - 6)}\\P(X = 7) = C(10, 7) * (0.632)^7 * (1 - 0.632)^{(10 - 7)}\\P(X = 8) = C(10, 8) * (0.632)^8 * (1 - 0.632)^{(10 - 8)}[/tex]
Now, let's calculate each of these probabilities:
[tex]P(X = 0) = 1 * 1 * (0.368)^{10} = 0.0010434068\\P(X = 1) = 10 * 0.632 * (0.368)^9 = 0.0091045617\\P(X = 2) = 45 * 0.632^2 * (0.368)^8 = 0.0333533058\\P(X = 3) = 120 * 0.632^3 * (0.368)^7 = 0.0789130084\\P(X = 4) = 210 * 0.632^4 * (0.368)^6 = 0.1312387057\\P(X = 5) = 252 * 0.632^5 * (0.368)^5 = 0.1552134487\\P(X = 6) = 210 * 0.632^6 * (0.368)^4 = 0.1320219902\\P(X = 7) = 120 * 0.632^7 * (0.368)^3 = 0.0702045343\\P(X = 8) = 45 * 0.632^8 * (0.368)^2 = 0.0272837154\\[/tex]
Now, let's sum up these probabilities:
P(X ≤ 8) ≈ 0.0010434068 + 0.0091045617 + 0.0333533058 + 0.0789130084 + 0.1312387057 + 0.1552134487 + 0.1320219902 + 0.0702045343 + 0.0272837154
≈ 0.6393716762
So, the probability of winning no more than 8 times out of 10 plays of the arcade game is approximately 0.6394.
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Finally, we find:
P(X ≤ 8) ≈ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
To calculate the probability of winning no more than 8 times out of 10 plays in the arcade game, we can use the binomial distribution.
The binomial distribution calculates the probability of having a certain number of successes (in this case, winning) in a fixed number of independent trials (10 plays) when the probability of success (p) remains constant.
Probability of winning (p) = 0.632
Probability of losing (q, since p + q = 1) = 0.368
Number of trials (n) = 10
To find the probability of winning no more than 8 times, we need to sum the probabilities of winning 0, 1, 2, 3, 4, 5, 6, 7, and 8 times.
P(X ≤ 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
Using the binomial probability formula:
P(X = k) = (n choose k) * p^k * q^(n-k)
where (n choose k) is the binomial coefficient, given by:
(n choose k) = n! / (k! * (n-k)!)
Calculating the probabilities for each individual case:
P(X = 0) = (10 choose 0) * (0.632^0) * (0.368^(10-0))
P(X = 1) = (10 choose 1) * (0.632^1) * (0.368^(10-1))
P(X = 2) = (10 choose 2) * (0.632^2) * (0.368^(10-2))
...
P(X = 8) = (10 choose 8) * (0.632^8) * (0.368^(10-8))
After calculating the probabilities for each individual case, we can sum them up to find P(X ≤ 8).
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