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In a newspaper printing factory, 10 out of every 100 newspapers are fo
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18 Aug 2023, 01:14
18 Aug 2023, 01:14
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Question Stats:
70% (01:51) correct
30% (01:23) wrong based on 10 sessions
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In a newspaper printing factory, 10 out of every 100 newspapers are found defective due to some defects in the printing. If two newspapers are randomly selected, what is the probability of getting exactly one defective newspaper?
A. 1/11
B. 2/11
C. 3/11
D. 4/11
E. 5/11
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A. 1/11
B. 2/11
C. 3/11
D. 4/11
E. 5/11
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
\(\Longrightarrow\) ALL the GRE Quant and Verbal Practice Directories in ONE place (2023)
\(\Longrightarrow\) Best GRE Test Books of 2023
Re: In a newspaper printing factory, 10 out of every 100 newspapers are fo
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11 Sep 2023, 00:39
11 Sep 2023, 00:39
Please explain this how the answer to this was derived.
What i did was using the generalised formula:
2C1(1/10)(9/10) =9/50
Posted from my mobile device
What i did was using the generalised formula:
2C1(1/10)(9/10) =9/50
Posted from my mobile device
In a newspaper printing factory, 10 out of every 100 newspapers are fo
[#permalink]
11 Sep 2023, 01:32
11 Sep 2023, 01:32
1
Expert Reply
OE
As we need to select two newspapers randomly out of the 100 newspaper.
Thus, The total number of ways of selecting two newspapers randomly =\( 𝐶^{100}_2\)
As 10 out of every 100 newspapers are found defective, thus there will be 90 non
defective newspapers.
Question wants to find exactly one defective newspaper, but we know that total
selection is of 2 newspapers.
Therefore, some has to be defective and one non-defective.
The probability will be
\(\dfrac{C^{10}_1+C^{90}_1}{C^{100}_2}=\dfrac{10 \times 90}{\frac{100 \times 99}{2}}=\frac{2}{11}\)
As we need to select two newspapers randomly out of the 100 newspaper.
Thus, The total number of ways of selecting two newspapers randomly =\( 𝐶^{100}_2\)
As 10 out of every 100 newspapers are found defective, thus there will be 90 non
defective newspapers.
Question wants to find exactly one defective newspaper, but we know that total
selection is of 2 newspapers.
Therefore, some has to be defective and one non-defective.
The probability will be
\(\dfrac{C^{10}_1+C^{90}_1}{C^{100}_2}=\dfrac{10 \times 90}{\frac{100 \times 99}{2}}=\frac{2}{11}\)
