At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 1/3 of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?
Correct answer: A
Explanation
2/3 of players who bat left-handed also bat right-handed.
Then 3/2(60%) = 90% of all players can bat left-handed.
P(a player does not bat left-handed) = 1 - 90% = 10%
The answer is A.
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AQuA-RAT items are crowdsourced algebra problems, not ACT questions. They carry five answer choices in the pre-2025 ACT Mathematics style and no ACT reporting category. Useful for drilling Preparing for Higher Math content; not a substitute for a real form.