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# Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,

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Joined: 02 Sep 2009
Posts: 52294
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

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29 Sep 2016, 03:35
1
1
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:58) correct 21% (01:44) wrong based on 125 sessions

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Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8

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Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

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29 Sep 2016, 03:48
1
Bunuel wrote:
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8

P(atleast ) = 1 - P(none)

Now we need to find none of the tv sets are broken.

We have 8 tvs all together..

First attempt only the fixed one = 6/8.

Second attempt only the fixed ones = 5/7...(since already one tv set is taken out)

both case = 6/8*5/7 = 15/28

P(atleast ) = 1 - P(none) => 1-15/28 = 13/28.

IMO option B
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Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

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29 Sep 2016, 04:34
total number of possible outcomes= 8C2= 28
Required is Total- (no of possibilities of selecting 2 repaired TVs)
28-6C2= 13
So probability= 13/28
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Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

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27 Feb 2017, 04:15
P(atleast one is broken) = 1- P(both fixed)
= 1 - 6C2/8C2
= 1- 15/28
= 13/28. Option B
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Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

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01 Mar 2017, 17:11
1
Bunuel wrote:
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8

We are given that there are 2 broken TV sets and 6 fixed TV sets and need to determine the probability that when 2 TV sets are selected, at least 1 is broken. Recall that the phrase “at least one” means “one or more.”

We can use the following formula:

1 = P(at least 1 set is broken) + P(no sets are broken)

Thus:

P(at least 1 set is broken) = 1 - P(no sets are broken)

P(no sets are broken) = 6/8 x 5/7 = 3/4 x 5/7 = 15/28

P(at least 1 set is broken) = 1 - 15/28 = 13/28

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Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

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08 Sep 2018, 07:46
Top Contributor
Bunuel wrote:
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8

There are 8 TV's in total
2 are broken
6 are fixed

We want to find P(at least one TV is broken)
When it comes to probability questions involving at least, it's often best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 broken TV) = 1 - P(not getting at least 1 broken TV)
What does it mean to not get at least 1 broken TV? It means getting zero broken TVs.

So, we can write: P(getting at least 1 broken TV) = 1 - P(getting zero broken TVs)
1 - P(getting two FIXED TVs)

P(getting two FIXED TVs)
P(getting two FIXED TVs) = P(1st TV fixed and 2nd TV is fixed)
= P(1st TV fixed) x P(2nd TV is fixed)
= 6/8 x 5/7
= 30/56
= 15/28

So, P(getting at least 1 broken TV) = 1 - P(not getting at least 1 broken TV)
= 1 - 15/28
= 13/28

Cheers,
Brent
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Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, &nbs [#permalink] 08 Sep 2018, 07:46
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