P(being replaced under warranty) = P(being replaced under warranty | submitted for service while under warranty) * P(submitted for service while under warranty)
= 0.4 * 0.25
= 0.1
n = 10
This is a binomial distribution, with n = 10 and p = 0.1
P(X = x) = 10Cx * 0.1x * (1 - 0.1)10-x
P(X = 2) = 10C2 * 0.12 * 0.98 = 0.1937
Question 2 0 out of 10 points Twenty-five percent of all telephones of a certain type...
Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones: What's the probability that at least 6 will end up being repaired under warranty?
Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly four will end up being replaced under warranty? (Round your answer to three decimal places.)
Thirty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly four will end up being replaced under warranty? (Round your answer to three decimal places.)
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Help me with the workings for these questions please!!!!
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