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At an elite baseball camp, 60% of players can bat both right-handed
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03 Jan 2015, 08:22
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At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
Re: At an elite baseball camp, 60% of players can bat both right-handed
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03 Jan 2015, 12:09
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Hi All,
This question can be considered an Overlapping Sets question (and solved with the Tic-Tac-Toe Board), but you don't need to organize your information in that way to get to the correct answer.
The key to this question is in realizing that some Left-handed batters ALSO bat with their Right-hands. Here's how you can answer the question by TESTing VALUES:
To start, let's choose 100 for the TOTAL number of players:
60% can bat with BOTH hands = 60 players
**Note: the TOTAL number of players who can bat with their LEFT hands include these 60 players AND the players who can bat with the Left hands ONLY.**
Next, we're told that 25% of players who can bat with their LEFT hands DO NOT bat with their Right hands.
X = Total who can bat Left-handed 25% of X DO NOT use their Right hands 75% of X CAN ALSO use their Right hands
Early on, we had 60 players who could bat with BOTH hands; THAT group is the 75% mentioned above...
.75X = 60 X = 80
There are 80 players who are Left-handed (20 are JUST Left-handed; 60 can bat with BOTH hands).
We're asked for the probability of randomly selecting a NON-Left-handed player.
At an elite baseball camp, 60% of players can bat both right-handed
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03 Jan 2015, 11:32
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whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
Let the total number of people who batted left handed = x total number of people who bat left-handed, but do not bat right handed = x/4 thus x= 60 + (x/4) or x=80
thus total number of people who batted right handed = 100-80 =20
Re: At an elite baseball camp, 60% of players can bat both right-handed
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03 Jan 2015, 19:54
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I felt the explanation in the file attached might make the task easy ..Correct me If Im wrong
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
Re: At an elite baseball camp, 60% of players can bat both right-handed
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02 Apr 2016, 16:16
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let L=total left handed batters L/4=left handed batters who do not bat right handed L-L/4=3L/4=left handed batters who also bat right handed let P=total players 3L/4=.6P P=5L/4 L/(5L/4)=4/5=80% probability that player selected at random will bat left handed 100%-80%=20% probability that player selected at random will not bat left handed
Re: At an elite baseball camp, 60% of players can bat both right-handed
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03 Jan 2015, 10:05
Apparently, I don't understand it either.
Three options:
Both = .6 Right only = ? Left only = .25
We are looking for Right only
1 - Both - Left only = .15
Right only = .15
Seem like the logical choice to me....
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
Re: At an elite baseball camp, 60% of players can bat both right-handed
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03 Jan 2015, 10:53
Me too failed to undstnd this ques.
am i missing something.
aeropower wrote:
Apparently, I don't understand it either.
Three options:
Both = .6 Right only = ? Left only = .25
We are looking for Right only
1 - Both - Left only = .15
Right only = .15
Seem like the logical choice to me....
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
Re: At an elite baseball camp, 60% of players can bat both right-handed
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01 Apr 2016, 23:07
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whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
I guess the tricky part is "25% of the players who bat left-handed do not bat right-handed". This also means that 75% of the players who bat left handed also bat right handed. => 75%of left handed players = 60. Therefore no. of left handed players = 80 Therefore number of players who bat only right handed = 20 Hence probability = 20/100 = 20%
Re: At an elite baseball camp, 60% of players can bat both right-handed
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01 Apr 2016, 23:42
MeghaP wrote:
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
I guess the tricky part is "25% of the players who bat left-handed do not bat right-handed". This also means that 75% of the players who bat left handed also bat right handed. => 75%of left handed players = 60. Therefore no. of left handed players = 80 Therefore number of players who bat only right handed = 20 Hence probability = 20/100 = 20%
Answer B
Please suggest if this approach is correct.
Hi, your approach is correct.. Mention that you are starting with the total number as 100
_________________
Re: At an elite baseball camp, 60% of players can bat both right-handed
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25 Nov 2019, 11:31
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whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% 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?
A.15% B. 20% C. 25% D. 30% E. 40%
We can solve using the Double Matrix Method.
The Double Matrix Method can be used for most questions featuring a population in which each member has two characteristics associated with it. Here, we have a population of baseball players, and the two characteristics are: - bats left-handed or DOESN'T bat left-handed - bats right-handed or DOESN'T bat right-handed
NOTICE that the question does not ask us to find an actual number. It asks us to find a probability. This means we can assign whatever value we wish to the total number of couples. So, let's say there are 100 players, which we'll add to our diagram:
60% of players can bat both right-handed and left-handed 60% of 100 = 60, so 60 players can bat both right-handed AND left-handed . Add that to the diagram to get:
25% of the players who bat left-handed do not bat right-handed Hmmm, we don't know the number of left-handed players, so we can't find 25% of that value. So, let's assign a variable. Let's let x = left-handed batters, and add it to our diagram: So, x of the 100 players bat left handed.
25% of the players who bat left-handed do not bat right-handed If x players bat left-handed, then 25% of x do not bat right-handed. In other words, 0.25x = number of players who do not bat right-handed Add this to our diagram:
At this point, we see that the two left-hand boxes add to x. So, we can write the equation: 60 + 0.25x = x Rearrange to get 60 = 0.75x Rewrite 0.75 as fraction to get: 60 = (3/4)x Multiply both sides by 4/3 to get: 80 = x If x = 80, then we know that 80 of the 100 players bat left-handed. This means that the remaining 20 players DO NOT bat left handed.
So, P(player doesn't bat left-handed) = 20/100 = 20%