M Notes Ask Your Teacher 4. -2 points Four salesmen play \"odd man out\" to see

Question

M Notes Ask Your Teacher 4. -2 points Four salesmen play "odd man out" to see who pays for lunch. They each flip a coin, and if there is a salesman whose coin doesn't match the others he pays for lunch. To clarify, the "odd man" must get heads while the other three get tails OR he must get tails while the other three get heads. Of course, since it is possible for two men get heads and two men get tails, not all flips will result in finding an "odd man". If this occurs the salesmen would be forced to flip again. What is the probability that there is an "odd man" the first time they flip?

Explanation / Answer

So, given there 4 sales man :

There are 4 sales man and each of the sales man can get either a Head or a tail we have (2^2)^4 outcomes in 1st attept = 256 outcomes

But, out of that HTTT, THTT, TTHT, TTTH,THHH, HTHH, HHTH, HHHT are just 8 outcomes that make some salesguy an 'oddman'
ending the game in 1st attempt

So, probability that there is an "odd man" for 1s time they flip is 8/256 = 1/32

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