Question

(c) In a game of chance the "attacker" and "defender" both roll one die. The attacker wins if his roll is larger. In the event of a tie, the defender wins. What is the probability the attacker wins? Ans. 0.4167] (d) In a lab class there are 5 males and 5 females. If partners are selected at random, what is the probability that all of the pairs have one male and one female? [Ans. 0.127]

Answer 1

(c)

If the defender rolls 1, then the attacker must roll 2, 3, 4, 5 or 6 to win. So in this case the probability that the attacker will win = (1/6) * (5/6). Here 1/6 is the probability that the defender will roll 1 and 5/6 is the probability that the attacker will roll any one of the 2, 3, 4, 5 or 6.

If the defender rolls 2, then the probability that attacker will win if he/she rolls 3, 4, 5 or 6 with probability = (1/6) * (4/6)

Same will go on for defender rolling 3, 4 and 5.

So the total probability = (1/6) * (5/6) + (1/6) * (4/6) + (1/6) * (3/6) + (1/6) * (2/6) + (1/6) * (1/6) = 15/36 = 0.4167

(d)

Basically, each person has 9 different other persons to choose from, and after that happens, the next pair's person has 7 other people to choose from (10 minus the first pair and himself) all the way down to the last person, that has only 1 other person to choose from.

so you have, N = 9 * 7 * 5 * 3 * 1 = (9)!! = 945

To form pair of only male and female each person (say male) will have 5 females to choose. After he chooses, the next person (Say male) will have 4 females to choose. After he chooses the next person (say female) will have 3 males to choose and it will go until the last male or female will choose the opposite gender.

So the number of ways are, P = 5 * 4 * 3 * 2 * 1 = 120

Desired probability = P/N = 120/945 = 0.127