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A certain store sells small, medium, and large toy trucks i

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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
[Reveal] Spoiler: OA

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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


let there be x number of trucks of all possible color-size combination.
probability that the truck she selects will have at least one of the two features Paul wants can be found by subtracting the probability of selecting a truck that doesn't has either of the property i.e. it is neither red in color nor of medim size from 1.

In the diagram attached, the I crossed off all the desired results i.e. cancelled all the possible combination of red-medium.
# of remaining outcomes= # of blue circles = # of trucks that don't have either of the desired property.

No. of colors=4
no. of sizes=3
Total outcomes=12

Remaining outcomes=6
hence probability that truck is neither red nor of medium size is \(6/12\).
hence probability that the truck she selects will have at least one of the two features Paul wants will be \(1 - 6/12\) or \(1/2\)
+1C
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probability.png
probability.png [ 9.42 KiB | Viewed 7364 times ]


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New post 08 Jan 2013, 07:53
Great Explanation! Thanks!
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]

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New post 08 Jan 2013, 09:05
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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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]

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New post 27 Jan 2013, 05:24
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


Please add a comma between blue and green. I mistook it as 1 category!
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]

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

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New post 08 Oct 2014, 17:05
I went with 1/2. I think the simplest way is to calculate 1 - P (0 features being chosen).
P (0 features) = 2/3 (the other two sizes) times 3/4 (the other two colors) = 1/2
1-1/2 = 1/2
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]

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New post 09 Oct 2014, 12:56
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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New post 10 Oct 2014, 02:33
Expert's post
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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New post 10 Oct 2014, 08:31
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

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New post 31 Mar 2015, 09:01
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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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


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).

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

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New post 18 Sep 2015, 16:53
can someone please explain why below approach is wrong.

assume there is one truck of each combo

Small 4 trucks ( 1 Red, 1 Blue, 1 Green, 1 yellow)

Medium 4 trucks ( 1 Red, 1 Blue, 1 Green, 1 yellow)

Large 4 trucks ( 1 Red, 1 Blue, 1 Green, 1 yellow)

Now, probability of choosing red truck is 3/12 = 1/4
probability of choosing medium truck is 4/12 = 1/3

probablity of either red or medium is 1/4 + 1/3 ( which is same as atleast red or medium)
= 7/12

This is obviously wrong answer but i cant figure where is the problem.
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Re: A certain store sells small, medium, and large toy trucks i [#permalink]

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New post 05 Apr 2016, 12:40
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 ,
can you show this by applyig P and C formula.

i tried the non event method, 1-3c2*4c3/3c1*4c1.This is coming out to be 0.There is some issue with the cases in the numerator of the fraction.
Re: A certain store sells small, medium, and large toy trucks i   [#permalink] 05 Apr 2016, 12:40
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