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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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Assume there are 1 of each giving - 3 sizes x 4 colours = 12 trucks

Since the q is asking for "at least one" (medium truck or red truck) let's calculate probability of selecting neither a red or medium truck and subtract from 1

Therefore - 1 - 6/12 = 1/2
Answer: C
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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The key is to take in account whenever see you the word \(AT\) \(Least\) is to subtract the other probability from 1

In this case you DO NO want the large and the small size so \(2/3\)\(AND\) (means \(*\)) the other colors so\(3/4\)

From this \(1 - 2/3 * 3/4 = 1 - 6/12 = 6/12 = 1/2\)
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
carcass wrote:
The key is to take in account whenever see you the word \(AT\) \(Least\) is to subtract the other probability from 1

In this case you DO NO want the large and the small size so \(2/3\)\(AND\) (means \(*\)) the other colors so\(3/4\)

From this \(1 - 2/3 * 3/4 = 1 - 6/12 = 6/12 = 1/2\)



I understand this reasoning and it makes perfect sense. But why in this case it doesn't work the probability of each:

1/3+1/4 = 7/12 ??

Thanks!
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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nachobioteck wrote:
carcass wrote:
The key is to take in account whenever see you the word \(AT\) \(Least\) is to subtract the other probability from 1

In this case you DO NO want the large and the small size so \(2/3\)\(AND\) (means \(*\)) the other colors so\(3/4\)

From this \(1 - 2/3 * 3/4 = 1 - 6/12 = 6/12 = 1/2\)



I understand this reasoning and it makes perfect sense. But why in this case it doesn't work the probability of each:

1/3+1/4 = 7/12 ??

Thanks!


You should subtract the overlap: 1/3*1/4 = 1/12 from 7/12.
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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let me explain in easy terms

the question asks atleast of the two(red-medium).So we gonna find trucks which dont have any of the mentioned or asked charateristic.i.e neither medium nor red and later subtract it.

first,total outcomes-4C1(for any colored truck)*3C1(size of the truck)=12
second,(truck neither medium nor red)=SB,SG,SY LB,LG,LY
so the required probality then toy selkected is medium-red or any of the two=1-(6/12)=1/2

other Solution=

either red or medium or both=MR,MB,MG,MY RS,RM,RL( HERE WE HAVE COUNTED MR or RM twice,so take only one)

prob=6/12=1/2

C.

hope it helps

-h
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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We have P(M or R) = P(M) + P(R) - P(M)*P(R) = 1/3 + 1/4 - 1/3*1/4 = 7/12 - 1/12 = 1/2.
M: Medium
R: Red
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A certain store sells small, medium, and large toy trucks i [#permalink]
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There are 3 sizes of trucks in 4 colours each, so a total of 3x4=12 trucks

P(M u R) = P(M) + P(R) - P(M n R)

=> 4/12 + 3/12 - 1/12
=> 6/12
=> 1/2 (Answer C)
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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fozzzy wrote:
A certain store sells small, medium, and large toy trucks in each of the colors red, blue, green, and yellow. The store has an equal number of trucks of each possible color-size combination. If Paul wants a medium, red truck and his mother will randomly select one the trucks in the store, what is the probability that the truck she selects will have at least one of the two features Paul wants?

A. 1/4
B. 1/3
C. 1/2
D. 7/12
E. 2/3

1. Favorable outcome is that, the truck should be medium or red or both
2. The opposite of this is easier to find which is neither medium nor red
3. Number of ways of selecting (large or small) and (blue , green or yellow) is 2*3=6
4. Total number of ways of selecting=3*4=12
5. Probability for opposite of favorable cases is 6/12=1/2
6. Probability for favorable case is 1-(1/2) = 1/2
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
Bunuel wrote:
nachobioteck wrote:
carcass wrote:
The key is to take in account whenever see you the word \(AT\) \(Least\) is to subtract the other probability from 1

In this case you DO NO want the large and the small size so \(2/3\)\(AND\) (means \(*\)) the other colors so\(3/4\)

From this \(1 - 2/3 * 3/4 = 1 - 6/12 = 6/12 = 1/2\)



I understand this reasoning and it makes perfect sense. But why in this case it doesn't work the probability of each:

1/3+1/4 = 7/12 ??

Thanks!


You should subtract the overlap: 1/3*1/4 = 1/12 from 7/12.


hi Bunuel , why should we subtract the overlap? since it says at least 1 of the features.shouldnt' it includes red, truck and red + truck?
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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jtankx90 wrote:
Bunuel wrote:
nachobioteck wrote:

I understand this reasoning and it makes perfect sense. But why in this case it doesn't work the probability of each:

1/3+1/4 = 7/12 ??

Thanks!


You should subtract the overlap: 1/3*1/4 = 1/12 from 7/12.


hi Bunuel , why should we subtract the overlap? since it says at least 1 of the features.shouldnt' it includes red, truck and red + truck?


hi...
overlap is added twice, ones with red and once with truck, so you have to subtract it once..
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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Hi All,

This question can be solved with a real simple "brute force" approach.

There are 3 types of toy trucks (small, medium and large) and 4 colors (red, blue, green and yellow). We're told that there are equal numbers of each color-size combination and we're asked what the probability is of randomly pulling a truck that is RED or MEDIUM or BOTH. I'm going to list out all of the possibilities and put a * next to the ones that fit what we're looking for:

Small Red *
Small Blue
Small Green
Small Yellow
Medium Red *
Medium Blue *
Medium Green *
Medium Yellow *
Large Red *
Large Blue
Large Green
Large Yellow

6 of the 12 options are red or medium or both.

Final Answer:

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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
Visual method:

Small, Medium, Large | Red, Green, Blue, Yellow |

4*3 =12

Child requires: Medium red trucks, or at least one.

\(Medium = M_{green};M_{red};M_{blue};M_{yellow}\)
\(Red=R_{small};R_{large}\)

Thus probability is
6 out of 12 -> \(\frac{6}{12}\) = \(\frac{1}{2}\)

Pay attention to not double count medium red trucks.
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
P (atleast one of two features) = 1 - P(neither of the feature)

P (not medium) = 2/3
P (not red) = 3/4

P (neither medium nor red) = 2/3 * 3/4 = 1/2

So, ans = P (atleast one of two features) = 1 - 1/2 = 1/2
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
OptimusPrepJanielle wrote:
fozzzy wrote:
A certain store sells small, medium, and large toy trucks in each of the colors red, blue, green, and yellow. The store has an equal number of trucks of each possible color-size combination. If Paul wants a medium, red truck and his mother will randomly select one the trucks in the store, what is the probability that the truck she selects will have at least one of the two features Paul wants?

A. 1/4
B. 1/3
C. 1/2
D. 7/12
E. 2/3


Probability of NOT selecting medium out of 3 sizes = 2/3
Probability of NOT selecting red out of 4 colours = 3/4

Total probability of NOT selecting red and medium = (2/3)*(3/4) = 1/2

Required probability = 1 - 1/2 (this will select at least one of red and medium)
= 1/2
Hence option (C).

--
Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course



VeritasKarishma

say if total there are 24 trucks

8 trucks per each size with 2 trucks for each color
probability of not choosing medium red truck is 16/24 i.e. 2/3

so 1-2/3 = 1/3 :? why its wrong ?


Also in the post above why are we multiplying events ? (2/3)*(3/4) = 1/2 is it because one can pick red and medium within one event ?
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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dave13 wrote:
OptimusPrepJanielle wrote:
fozzzy wrote:
A certain store sells small, medium, and large toy trucks in each of the colors red, blue, green, and yellow. The store has an equal number of trucks of each possible color-size combination. If Paul wants a medium, red truck and his mother will randomly select one the trucks in the store, what is the probability that the truck she selects will have at least one of the two features Paul wants?

A. 1/4
B. 1/3
C. 1/2
D. 7/12
E. 2/3


Probability of NOT selecting medium out of 3 sizes = 2/3
Probability of NOT selecting red out of 4 colours = 3/4

Total probability of NOT selecting red and medium = (2/3)*(3/4) = 1/2

Required probability = 1 - 1/2 (this will select at least one of red and medium)
= 1/2
Hence option (C).

--
Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course



VeritasKarishma

say if total there are 24 trucks

8 trucks per each size with 2 trucks for each color
probability of not choosing medium red truck is 16/24 i.e. 2/3

so 1-2/3 = 1/3 :? why its wrong ?


Also in the post above why are we multiplying events ? (2/3)*(3/4) = 1/2 is it because one can pick red and medium within one event ?


Hi dave13,

In my explanation (above your post), I presented an example that included 12 trucks. We can easily modify it for 24 trucks though...

There are 3 types of toy trucks (small, medium and large) and 4 colors (red, blue, green and yellow). We're told that there are equal numbers of each color-size combination and we're asked what the probability is of randomly pulling a truck that is RED or MEDIUM or BOTH. Assuming that the TOTAL number of trucks is 24, here is a list of all of the possibilities (and I've put a * next to the ones that fit what we're looking for):

2 Small Red *
2 Small Blue
2 Small Green
2 Small Yellow
2 Medium Red *
2 Medium Blue *
2 Medium Green *
2 Medium Yellow *
2 Large Red *
2 Large Blue
2 Large Green
2 Large Yellow

12 of the 24 options are red or medium or both. So 1/2 of the options fit what we are asked for and 1/2 of the trucks do not.

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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
Size Colour Qty (Nos)
Small Red 1
Blue 1
Green 1
Yellow 1
4
Medium Red 1
Blue 1
Green 1
Yellow 1
4
Large Red 1
Blue 1
Green 1
Yellow 1
4

Paul wants the ball (medium red - MR)
One ball selected in which at least one feature is taken care of - SR, MR, MB, MG, MY, LR
No of possible cases - 6
Total possible selections 12
Probability :1/2
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]
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dave13 wrote:
OptimusPrepJanielle wrote:
fozzzy wrote:
A certain store sells small, medium, and large toy trucks in each of the colors red, blue, green, and yellow. The store has an equal number of trucks of each possible color-size combination. If Paul wants a medium, red truck and his mother will randomly select one the trucks in the store, what is the probability that the truck she selects will have at least one of the two features Paul wants?

A. 1/4
B. 1/3
C. 1/2
D. 7/12
E. 2/3


Probability of NOT selecting medium out of 3 sizes = 2/3
Probability of NOT selecting red out of 4 colours = 3/4

Total probability of NOT selecting red and medium = (2/3)*(3/4) = 1/2

Required probability = 1 - 1/2 (this will select at least one of red and medium)
= 1/2
Hence option (C).

--
Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course



VeritasKarishma

say if total there are 24 trucks

8 trucks per each size with 2 trucks for each color
probability of not choosing medium red truck is 16/24 i.e. 2/3

so 1-2/3 = 1/3 :? why its wrong ?


Also in the post above why are we multiplying events ? (2/3)*(3/4) = 1/2 is it because one can pick red and medium within one event ?


If there are 24 trucks, 8 will be small (2 red) , 8 medium (2 red) and 8 large (2 red).

No. of trucks that have at least one desirable characteristic = 2 (red but small) + 8 (all medium) + 2 (red but large)
All these 12 trucks have at least one of the 2 features Paul wants.

So out of 24, there are 12 trucks that have at least one desirable feature. Probability = 12/24 = 1/2
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