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

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AAPL: Moderator; Posts: 2632; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Sat Sep 08, 2018 5:13 am

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Economist GMAT

Out if 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

OA B.

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Source: Beat The GMAT — Problem Solving | Sat Sep 08, 2018 5:13 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Sep 08, 2018 7:41 am

AAPL wrote:Economist GMAT

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

OA B.

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

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Sat Sep 08, 2018 2:05 pm

AAPL wrote:Economist GMAT

Out if 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

\[\left. \begin{gathered}
2\,{\text{b}}\,\,\left( {{\text{broken}}} \right)\,\, \hfill \\
6\,{\text{f}}\,\,\,\left( {{\text{fixed}}} \right) \hfill \\
\end{gathered} \right\}\,\,\,2\,\,{\text{chosen}}\,\,{\text{simultaneously}}\,\,\,\left( {{\text{randomly}}} \right)\]
\[? = P\left( { \geqslant 1\,{\text{b}}} \right) = 1 - P\left( {2\,{\text{f}}} \right)\]
\[P\left( {2\,{\text{f}}} \right) = \frac{{C\left( {6,2} \right)}}{{C\left( {8,2} \right)}} = \ldots = \frac{{15}}{{28}}\]
\[? = 1 - \frac{{15}}{{28}} = \boxed{\frac{{13}}{{28}}}\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8136; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Mon Sep 17, 2018 5:35 pm

AAPL wrote:Economist GMAT

Out if 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

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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