Scenario: Eric and Joshua are playing ping pong and pool. Joshua believes he has a good

chance of beating Eric in at least one of the games. The probability Joshua beats Eric in

ping pong is 0.48. The probability Joshua beats Eric in pool is 0.46. Joshua is willing to

assume the probability of Eric winning a game of ping pong is independent of him winning

a game of pool.

Round all answers to TWO decimal places

The probability that Joshua beats Eric in ping pong AND pool is

The probability that Joshua beats Eric in ping pong OR pool is

a) 0.22

b) 0.50

Step-by-step 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)