The probability that Richard beats John at badminto is 0.7. The probability that Richard beats John at sqaush is 0.6. These events are independant. Calculate the probabilty that, in a week when they play one game of badminton and sqaush:
a) Richard wins both games
b) Richard wins one game and loses the other.
a) .7 * .6 = ?
b) Which one does he lose?
Loses badminton = .3 * .6 = ?
Loses squash = .7 * .4 = ?
For "either-or" probability among the two above, add the newly found probabilities.
To calculate the probability of these events happening, we can multiply the probabilities of each individual event.
a) Probability of Richard winning both games:
The probability of Richard winning at badminton is 0.7.
The probability of Richard winning at squash is 0.6.
Since the events are independent, we can multiply the probabilities together:
0.7 * 0.6 = 0.42
Therefore, the probability of Richard winning both games is 0.42.
b) Probability of Richard winning one game and losing the other:
To calculate this probability, we need to consider the two possible scenarios:
i) Richard wins at badminton and loses at squash:
The probability of Richard winning at badminton is 0.7.
The probability of Richard losing at squash is 1 - 0.6 = 0.4
Since the events are independent, we can multiply the probabilities together:
0.7 * 0.4 = 0.28
ii) Richard loses at badminton and wins at squash:
The probability of Richard losing at badminton is 1 - 0.7 = 0.3
The probability of Richard winning at squash is 0.6.
Since the events are independent, we can multiply the probabilities together:
0.3 * 0.6 = 0.18
To find the total probability, we can add the probabilities of the two scenarios:
0.28 + 0.18 = 0.46
Therefore, the probability of Richard winning one game and losing the other is 0.46.
To calculate the probabilities, we can multiply the individual probabilities together for the first scenario, and use a combination of addition and multiplication for the second scenario.
Let's start by calculating the probability that Richard wins both games:
a) Richard wins both games:
Since the events (Richard winning at badminton and Richard winning at squash) are independent, we can multiply the probabilities:
Probability of Richard winning at badminton = 0.7
Probability of Richard winning at squash = 0.6
Probability of Richard winning both games = 0.7 * 0.6 = 0.42
So, the probability that Richard wins both games is 0.42.
Now, let's calculate the probability that Richard wins one game and loses the other:
b) Richard wins one game and loses the other:
To calculate this probability, we need to consider two separate scenarios:
1. Richard wins at badminton and loses at squash.
2. Richard loses at badminton and wins at squash.
Scenario 1:
Probability of Richard winning at badminton = 0.7
Probability of Richard losing at squash = 1 - 0.6 (since losing is the complement of winning)
Probability of Richard winning at badminton and losing at squash = 0.7 * (1 - 0.6) = 0.7 * 0.4 = 0.28
Scenario 2:
Probability of Richard losing at badminton = 1 - 0.7 (since losing is the complement of winning)
Probability of Richard winning at squash = 0.6
Probability of Richard losing at badminton and winning at squash = (1 - 0.7) * 0.6 = 0.3 * 0.6 = 0.18
To calculate the total probability, we sum up the probabilities from both scenarios:
Probability of Richard winning one game and losing the other = Scenario 1 + Scenario 2 = 0.28 + 0.18 = 0.46
So, the probability that Richard wins one game and loses the other is 0.46.
To summarize:
a) The probability that Richard wins both games is 0.42.
b) The probability that Richard wins one game and loses the other is 0.46.