Question

In the game of "odd man out" each player tosses a fair coin. if all the coins turn up the same except for one, the player tossing the different coin is declared the odd man out and is eliminated from the contest. suppose that three people are playing. what is the probability that someone will be eliminated on the first toss?

Answer 1

The probability that someone will be eliminated on the first toss in the game of "odd man out" is 3/4.

Step-by-step explanation:

To find the probability that someone will be eliminated on the first toss in the game of "odd man out," we need to consider the possible outcomes.

There are 8 possible outcomes when three coins are tossed: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT.

Out of these outcomes, 6 of them have one coin that is different from the others (HHT, HTH, THH, TTH, THT, HTT).

Therefore, the probability of someone being eliminated on the first toss is:

P(elimination on first toss) = Number of favorable outcomes / Total number of outcomes = 6 / 8 = 3 / 4.

Answer 2

The options are H,H,T or T,T,H

HHT: 1/2 x 1/2 x 1/2 = 1/8

TTH: 1/2 x 1/2 x 1/2 = 1/8

HHT or TTH: 1/8 + 1/8 = 2/8 = 1/4

Answer: 1 out of 4 = 1/4 = 25%