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 alike, 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 50,000,000 k. 10^7 (10^7 - 1)/2k 50,000,000/k^2. 10^7 (10^7 -1)/2k^2 10,000,000/k

Explanation / Answer

here as each car key has probability that it belogs to a random car p=1/k

total number of cases in one year n =5*107

hence expected number of cases =np=50000000/K

option A is correct

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