Answer:
a) 0.22
b) 0.50
Explanation
Solution:-
- Let Event ( A ) be defined as Joshua defeats Eric in a game of ping-pong. The probability for event ( A ) is given as:
p ( A ) = 0.48
- Then the probability for Eric winning the ping-pong match against Joshua would be:
p ( A' ) = 1 - p ( A )
p ( A' ) = 1 - 0.48
p ( A' ) = 0.52
- Let Event ( B ) be defined as Joshua defeats Eric in a game of pool. The probability for event ( B ) is given as:
p ( B ) = 0.46
- Similarly, the probability for Eric winning the pool match against Joshua would be:
p ( B' ) = 1 - p ( B )
p ( B' ) = 1 - 0.46
p ( B' ) = 0.54
- The probability of Joshua or Eric winning either of the games are independent from each other.
- We are to determine the probability that Joshua beats Eric in ping pong AND pool.
- We are looking for a probability of events ( A & B ) to be true. Since the probability of Joshua winning either of the game is independent from each other. Therefore, use the property of independent events:
p ( A & B ) = p ( A ) * p ( B )
p ( A & B ) = 0.48 * 0.46
Answer: p ( A & B ) = 0.2208 ... ( 0.22 rounded to 2 dp )
- We are to determine the probability that Joshua beats Eric in ping pong OR pool.
- We are looking for a probability of either of the events ( A or B ) to be true. Since the probability of Joshua winning either of the game is independent from each other:
- We have two cases that Joshua wins ping-pong match and loose pool OR loose ping-pong game and win the pool match against Eric. Hence,
p ( A U B ) = p ( A ) * p ( B' ) + p ( A' ) * p ( B )
p ( A U B ) = 0.48 * 0.54 + 0.52 * 0.46
Answer: p ( A U B ) = 0.4984 ... ( 0.50 rounded to 2 dp )
Alternate Method part ( b ):
- We can also use the property of event sets to determine the probability of Union of two sets/events ( A and B ):
P ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )
= 0.48 + 0.46 - 2*0.2208
Answer: p ( A U B ) = 0.4984 ... ( 0.50 rounded to 2 dp )