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At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
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whitehalo
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%

Answer: B

Cheers,
Brent

RELATED VIDEO

Originally posted by BrentGMATPrepNow on 25 Nov 2019, 11:31.
Last edited by BrentGMATPrepNow on 17 May 2021, 09:52, edited 1 time in total.
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
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
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?
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
1
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Me too failed to undstnd this ques.

am i missing something.

aeropower
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
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?
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
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I felt the explanation in the file attached might make the task easy ..Correct me If Im wrong


whitehalo
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?

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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
1
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whitehalo
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.
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
Expert Reply
MeghaP
whitehalo
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
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
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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
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
The trick is that it's a set problem and not a venn diagram.

So if 25% of the batters who bat left don't bat right that means it's 20%. As if 20 people don't bat right and 60% do then that's 2/8 or 25%
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
Expert Reply
whitehalo
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%


Solution:

We can let the total number of players = 100. So we have 60 players who can bat both right-handed and left-handed. We can let x = the number of players who can only bat left-handed. Thus, 40 - x = the number of players who can only bat right handed. We are given that 25% of the players who bat left-handed do not bat right-handed, which means they can only bat left-handed. Therefore, we can create the equation:

0.25(x + 60) = x

x + 60 = 4x

60 = 3x

20 = x

So we have 20 players who can only bat left-handed and also 40 - 20 = 20 players who can only bat right handed. We are asked for the probability that a player selected at random does not bat left-handed, i.e., who can only bat right-handed. Therefore, that probability is 20/100 = 20%.

Answer: B
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
60% R/L
25% of all L is NR/L

60+1/4x=x
60=3/4x
x=80

100-80=20
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
BrentGMATPrepNow
whitehalo
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%

Answer: B

Cheers,
Brent

RELATED VIDEO

Could you explain how we can use NOT left and NOT right? Isn't NOT left = Right and NOT right = Left?
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
Expert Reply
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Hoozan

Could you explain how we can use NOT left and NOT right? Isn't NOT left = Right and NOT right = Left?

I used those terms in order to distinguish the four different categories of players:
1) bats left handed but not right handed
2) bats right handed but not left handed
3) bats left handed and right handed
4) bats neither left handed nor right handed
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
Assume 100 people

60% of the 100 people = 60 ppl are (L & R)


Assume L = total number of people who had Left Handed ——-> L will include:


Players who are (L ONLY)
+
Players who bat (L & R)


L = (L Only) + (L & R)

L = (L Only) + (60)

“25% of the batters who bat LEFT HANDED do not bat right handed”

Translated: 25% of the L total are part of (L Only)

This means that (100 - 25)% of the L total must bat BOTH (L & R)


75% * L = Both (L & R) = 60

(3/4)L = 60

L = 80

Out of the 100 people, 80 people bat Left Handed

The amount of people who do NOT bat Left Handed or (R Only) must be the remaining 20 people


Probability = 20 / 100


20 %

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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
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Re: At an elite baseball camp, 60% of players can bat both right-handed [#permalink]
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