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# Mike's pencil box contains 30 pens. 15 of the pens are red, and all of

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Mike's pencil box contains 30 pens. 15 of the pens are red, and all of  [#permalink]

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Updated on: 24 May 2015, 06:52
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75% (01:20) correct 25% (01:09) wrong based on 203 sessions

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Mike's pencil box contains 30 pens. 15 of the pens are red, and all of the others are either blue or black. If Mike were to choose a random pen from the pencil box, what is the probability that it would be blue?

(1) The probability that the pen will be red minus the probability that the pen will be black equals 0.3
(2) The probability that the pen will be red or blue is 0.8

Could anyone help me calculating the probabilities of picking a blue pencil? DS is clear for me, but the true calculation of the numbers...

According to Stat.(1), The probability that the pen will be red minus the probability that the pen will be black equals 0.3. This would allow us to calculate the probability of the pen being black. (since P(red)-P(black)=0.3.).

If you know the probability of black and red, you know the probability of blue:

P(red)+P(black)+P(blue) = 1

Therefore this statement is sufficient. Stat.(1)->S->AD.

According to Stat.(2), The probability that the pen will be red or blue is 0.8. Since the probability of a single event with two possible outcomes with an OR relation equals to the sum of their probabilities, this means that:

P(red)+P(blue)= 0.8

By plugging in P(red) in the equation above, it is possible to calculate P(blue). Therefore, this statement is sufficient. Stat.(2)->S->D.

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Originally posted by reto on 24 May 2015, 05:47.
Last edited by Bunuel on 24 May 2015, 06:52, edited 1 time in total.
Edited the question.
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Re: Mike's pencil box contains 30 pens. 15 of the pens are red, and all of  [#permalink]

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24 May 2015, 06:26
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Hi reto,

We are given in the question the number of red pens and the total number of pens. Hence we can find the probability of a red pen being picked up.

Total number of pens = 30
Total number of red pens = 15

P(Red Pen) = $$\frac{15}{30} = \frac{1}{2} = 0.5$$

Probability of a black or a blue pen being picked up is directly dependent on the number of black and blue pens respectively.

Statement-I
St-I tells us that P(Red) - P(Black) = 0.3
0.5 - P(Black) = 0.3 i.e. P(Black) = 0.2. We can interpret this as the number of black pens being 20% of the total pens.

So Number of black pens = 20% of 30 = 6. From here we can say that total number of blue pens = 30 - 15 - 6 = 9

Statement-II
St-II tells us that P(Red or Blue) = 0.8 i.e. P(Red) + P(Blue) - P(Red and Blue) = 0.8.

Since a pen can't be red and blue at the same time, therefore P(Red and Blue) = 0.
P(Red) + P(Blue) = 0.8

0.5 + P(Blue) = 0.8 i.e. P(Blue) = 0.3.We can interpret this as the number of blue pens being 30% of the total pens.

So Number of blue pens = 30% of 30 = 9. From here we can say that total number of black pens = 30 - 15 - 9 = 6.

Hope this helps

Regards
Harsh
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Re: Mike's pencil box contains 30 pens. 15 of the pens are red, and all of  [#permalink]

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24 May 2015, 09:25
reto wrote:
Mike's pencil box contains 30 pens. 15 of the pens are red, and all of the others are either blue or black. If Mike were to choose a random pen from the pencil box, what is the probability that it would be blue?

(1) The probability that the pen will be red minus the probability that the pen will be black equals 0.3
(2) The probability that the pen will be red or blue is 0.8

Could anyone help me calculating the probabilities of picking a blue pencil? DS is clear for me, but the true calculation of the numbers...

When you pick just one thing from a group, probabilities are just proportions. So here, the probability of picking a red pen is 15/30 = 1/2, or in other words, 1/2 the pens are red, or 50% of the pens are red. So it's not really a probability question at all; it's just a ratio question.

We have a box where 50% of our pens are red, and the rest are blue and black. Statement 2 is slightly easier - it tells us that 80% are either red or blue, so if 50% are red, then 30% are blue. Similarly, Statement 1 tells us that 20% of the pens are black, so 30% will be blue.
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Re: Mike's pencil box contains 30 pens. 15 of the pens are red, and all of  [#permalink]

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27 Jun 2018, 18:14
reto wrote:
Mike's pencil box contains 30 pens. 15 of the pens are red, and all of the others are either blue or black. If Mike were to choose a random pen from the pencil box, what is the probability that it would be blue?

(1) The probability that the pen will be red minus the probability that the pen will be black equals 0.3
(2) The probability that the pen will be red or blue is 0.8

We see that the P(red) = 15/30 = 0.5 and thus P(blue or black) = P(blue) + P(black) = 0.5. We need to determine P(blue). So if we know P(black), then we can determine P(blue) since P(blue) = 0.5 - P(black).

Statement One Alone:

The probability that the pen will be red minus the probability that the pen will be black equals 0.3.

So we have:

P(red) - P(black) = 0.3

0.5 - P(black) = 0.3

P(black) = 0.2

Thus, P(blue) = 0.5 - 0.2 = 0.3. Statement one alone is sufficient.

Statement Two Alone:

The probability that the pen will be red or blue is 0.8.

So we have:

P(red) + P(blue) = 0.8

0.5 + P(blue) = 0.8

P(blue) = 0.3

Statement two alone is also sufficient.

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Re: Mike's pencil box contains 30 pens. 15 of the pens are red, and all of &nbs [#permalink] 27 Jun 2018, 18:14
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