Answer:
The probability that a player selected at random does not bat left-handed is 20%.
Explanation:
Assume that there are a total of 100 baseball players at an elite baseball camp.
The information provided is:
- 60% of players can bat both right-handed and left-handed.
- 25% of the players who bat left-handed do not bat right-handed.
That is, the number of players can bat both right-handed and left-handed is,
n (L and R) = 60.
Now, if 25% of the players who bat left-handed do not bat right-handed, then 75% of all left-handed players can also bat right-handed.
⇒ n (L and R) = n (L) × 75%
60 = n (L) × 0.75
n (L) = 60/0.75
n (L) = 80
So there are 80 left handed batters.
Compute the number of only left handed batters as follows:
n (Only L) = n (L) × 25%
= 80 × 0.25
= 20
So there are 20 only left handed batters.
Compute the number of only right handed batters as follows:
Total = n (Only L) + n (Only R) + n (L and R)
100 = 20 + n (Only R) + 60
n (Only R) = 20
Thus, the probability that a player selected at random does not bat left-handed is 20%.