Three boys play a game of luck in which their respective chances of winning are 1/2 1/3 and 1/4.What is the probability that one and only one of the boys wins the game?
Pr(first boy) = A = 1/2
Pr(second boy) = B = 1/3
Pr(third boy) = C = 1/4
A, B and C can win = 1/2*1/3*1/4
=1/24
A wins, B wins, C looses=1/2*1/3*3/4
3/24
A wins, B looses, C wins=1/2*2/3*1/4
2/24
A looses, B wins , C wins=1/2*1/3*1/4
1/24
Probability at least one of them win=1/24 +3/24 + 2/24 + 1/24
=1/24 +3/24 + 2/24 + 1/24
=7/24
To find the probability that one and only one of the boys wins the game, we need to consider all possible scenarios where one boy wins and the other two boys lose.
Let's calculate the probability for each boy winning and the others losing:
Boy 1 wins, Boys 2 and 3 lose:
Probability of Boy 1 winning = 1/2
Probability of Boy 2 losing = 2/3 (since the probability of winning for Boy 2 is 1/3)
Probability of Boy 3 losing = 3/4 (since the probability of winning for Boy 3 is 1/4)
So, the probability of this scenario is (1/2) * (2/3) * (3/4) = 1/4
Similarly, we calculate the other two scenarios:
Boy 2 wins, Boys 1 and 3 lose:
Probability of Boy 1 losing = 1/2
Probability of Boy 2 winning = 1/3
Probability of Boy 3 losing = 3/4
Probability of this scenario = (1/2) * (1/3) * (3/4) = 1/8
Boy 3 wins, Boys 1 and 2 lose:
Probability of Boy 1 losing = 1/2
Probability of Boy 2 losing = 2/3
Probability of Boy 3 winning = 1/4
Probability of this scenario = (1/2) * (2/3) * (1/4) = 1/12
Finally, we add up the probabilities of each scenario to get the total probability:
Total probability = Probability of Boy 1 winning and the others losing + Probability of Boy 2 winning and the others losing + Probability of Boy 3 winning and the others losing
= 1/4 + 1/8 + 1/12
= 3/12 + 2/12 + 1/12
= 6/12
= 1/2
Therefore, the probability that one and only one of the boys wins the game is 1/2.
To find the probability that one and only one of the boys wins the game, we need to consider three possible scenarios: Boy 1 wins, Boy 2 wins, or Boy 3 wins.
First, let's calculate the probability that Boy 1 wins and the other two boys don't win. The probability of Boy 1 winning is 1/2, and the probability of Boy 2 not winning is 2/3 (since they have a 1/3 chance each). Likewise, the probability of Boy 3 not winning is 3/4. Since these events are independent, we can multiply their probabilities together:
P(Boy 1 wins and Boy 2 doesn't win and Boy 3 doesn't win) = (1/2) * (2/3) * (3/4)
Next, let's calculate the probability that Boy 2 wins and the other two boys don't win. The probability of Boy 2 winning is 1/3, and the probability of Boy 1 not winning is 1/2, while the probability of Boy 3 not winning is 3/4:
P(Boy 2 wins and Boy 1 doesn't win and Boy 3 doesn't win) = (1/3) * (1/2) * (3/4)
Finally, let's calculate the probability that Boy 3 wins and the other two boys don't win. The probability of Boy 3 winning is 1/4, and the probability of Boy 1 not winning is 1/2, while the probability of Boy 2 not winning is 2/3:
P(Boy 3 wins and Boy 1 doesn't win and Boy 2 doesn't win) = (1/4) * (1/2) * (2/3)
Now, we simply add up the probabilities of these three exclusive events to find the probability that one and only one of the boys wins the game:
P(One and only one of the boys wins) = P(Boy 1 wins and Boy 2 doesn't win and Boy 3 doesn't win) + P(Boy 2 wins and Boy 1 doesn't win and Boy 3 doesn't win) + P(Boy 3 wins and Boy 1 doesn't win and Boy 2 doesn't win)
= (1/2) * (2/3) * (3/4) + (1/3) * (1/2) * (3/4) + (1/4) * (1/2) * (2/3)
Calculating these values gives us the desired probability.