In a drawer of shirts 8 are blue, 6 are green and 4 are

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 am; Followed by:3 members

In a drawer of shirts 8 are blue, 6 are green and 4 are

by VJesus12 » Fri May 11, 2018 7:36 am

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In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

(A) 25/153
(B) 28/153
(C) 5/17
(D) 4/9
(E) 12/17

The OA is the E.

I am really confused here. Can anyone give me some help, please? Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Fri May 11, 2018 7:36 am

Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Fri May 11, 2018 8:54 am

VJesus12 wrote:In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

(A) 25/153
(B) 28/153
(C) 5/17
(D) 4/9
(E) 12/17

The OA is the E.

I am really confused here. Can anyone give me some help, please? Thanks in advance.

Hello vjesus12.

We have 18 colors.

Since we want to find the probability of at least one of the shirts he draws will be blue, we can find it as follows: $$P\left(at\ least\ 1\ blue\right)=1-P\left(no\ blue\right)$$ Now, the probability of selecting no blues is: $$P\left(no\ blue\right)=\frac{10}{18}\cdot\frac{9}{17}=\frac{5}{9}\cdot\frac{9}{17}=\frac{5}{17}.$$ Hence we get: $$P\left(at\ least\ 1\ blue\right)=1-\frac{5}{17}=\frac{12}{17}.$$ Therefore, the correct answer is the option E.

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nidhips: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Tue May 01, 2018 4:55 pm

by nidhips » Fri May 18, 2018 3:14 pm

My solution was to calculate 1- 10c 2/18c2 = 108/153=12/17

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8087; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue May 22, 2018 5:15 pm

VJesus12 wrote:In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

(A) 25/153
(B) 28/153
(C) 5/17
(D) 4/9
(E) 12/17

We can use the equation:

P(at least one of the shirts he draws will be blue) = 1 - P(no blue shirts)

P(no blue shirts) = 10/18 x 9/17 = 5/9 x 9/17 = 5/17.

So, P(at least one of the shirts he draws will be blue) = 1 - 5/17 = 12/17.

Answer: E

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