Suppose that 3 people toss an unfair coin that has a probabi

Suppose that 3 people toss an unfair coin that has a probability of 1/4 of landing on a head. The person who is the odd one out (their coin lands on a different side) pays for lunch. If all coins turn up the same, they are tossed again. Find the probability that fewer than 4 tossess are needed to find the person who pays for lunch.

Solution

We want to find the probability that fewer than 4 tosses are needed

One toss is needed , this mean one person got the odd

HHT

HTH

THH

TTH

THT

HTT

There are 6 out of total 8 cases in which one man gets the odd, and pays for the lunch

P(one toss) = 6/8

.

P(two tosses are needed) = this is done when the first toss had all same sides (HHH or TTT) , thus there are 2 out of 8 outcomes that give same faces P(first toss) = 2/8 , and the second toss tells us who needs to pay P(second toss) = 6/8

Thus P(two tosses) = 2/8 * 6/8

Likewise P(three tosses) = 2/8 * 2/8 * 6/8

.

THus P(fewer than 4 tosses) = P(one toss) + P(two tosses) + P(three tosses)

P(fewer than 4 tosses) = 6/8 + 2/8*6/8 + 2/8 * 2/8 * 6/8

P(fewer than 4 tosses) = 3/4 + 3/16 + 3/64

P(fewer than 4 tosses) = 3/4 ( 1 + 1/4 + 1/16)

P(fewer than 4 tosses) = 0.984375