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22 Mar 2018, 08:15
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Samuel has a big drawer full of exactly 59 packets, each containing a marker which is either black or blue or red in colour. All 59 packets are sealed and are completely identical from outside. It is not possible to know which colour marker is inside unless the packet is fully opened. Out of 59, 23 packets contain a black marker each. How many packets contain a blue marker?
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
22 Mar 2018, 09:23
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TOTAL = 59
Black = 23
Blue = x
Red = y
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
In that case, 23+ 18 + 1 = 42 and he has at least one marker of each color. Then Black = 23, Blue = 18, Red = 18. SUFFICIENT
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
In that case R/59 = Blue/59 --> R = B. Then 59 - 23 (Black) = 36.
As Blue = Red, 36/2 = 18 = Red = Blue. SUFFICIENT
Ans: (D)
Please correct me if I'm wrong
Black = 23
Blue = x
Red = y
(1) If Samuel withdraws packets without looking at their contents, he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
In that case, 23+ 18 + 1 = 42 and he has at least one marker of each color. Then Black = 23, Blue = 18, Red = 18. SUFFICIENT
(2) Probability of picking a packet containing red marker is same as the probability of picking a packet containing blue marker.
In that case R/59 = Blue/59 --> R = B. Then 59 - 23 (Black) = 36.
As Blue = Red, 36/2 = 18 = Red = Blue. SUFFICIENT
Ans: (D)
Please correct me if I'm wrong
22 May 2018, 18:40
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Given:
59 total
23 black
36 blue or red
Find blue.
Statement 1
42 packets ensures that you'll have one of each color. This means that in any 42 packets you withdraw, you'll have:
- ALL of the packets of the color that has the most packets,
- ALL of the packets of the color that has the second most packets, and
- ONE of the packets of the color that has the least packets.
In any selection of 42 packets, 23 will be black, and 19 will be blue or red... so black has the most packets (23), and either blue or red has the second most (18). But note that in our initial analysis, blue+red=36. Thus, if one of the colors has 18 packets, the other will also have 18 packets.
Sufficient
Statement 2
36 packets of blue or red, half of them are red, thus half of them are blue.
Sufficient
Answer: D
59 total
23 black
36 blue or red
Find blue.
Statement 1
42 packets ensures that you'll have one of each color. This means that in any 42 packets you withdraw, you'll have:
- ALL of the packets of the color that has the most packets,
- ALL of the packets of the color that has the second most packets, and
- ONE of the packets of the color that has the least packets.
In any selection of 42 packets, 23 will be black, and 19 will be blue or red... so black has the most packets (23), and either blue or red has the second most (18). But note that in our initial analysis, blue+red=36. Thus, if one of the colors has 18 packets, the other will also have 18 packets.
Sufficient
Statement 2
36 packets of blue or red, half of them are red, thus half of them are blue.
Sufficient
Answer: D
