Suppose I am a car manufacturer. Rather than actually making all the ignition ke

Question

Suppose I am a car manufacturer. Rather than actually making all the ignition keys unique, l decide to make just k different kinds of keys. Each new car gets assigned one of these kinds at random, independently of how other cars' keys are assigned. Now suppose there are 10 million cars of a particular model manufactured. Because the cars all look a like, car owners occasionally try to unlock the doors of other people's cars by accident. When they succeed, it inevitably makes the next day's newspaper. If each car owner tries to open 5 other people's cars during the course of a year, what is the expected number of newspaper reports (that is, the expected number of times some car owner will open another's car? You can assume that the attempt will succeed if and only if the two cars have the same kind of key. 10^7(10^7 - 1)/2k 50,000,000/k 50,000,000/k^2. 10^7(10^7 - 1)/2k^2 10,000,000/k

Explanation / Answer

total number of keys =1/k

hence probabilty that a key beongs to a random car p=1/k

total number of cases n=5*10000000=50000000

hence expectation =E(X) =np =50000000/k

therefore option B is correct

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