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A certain machine produces toy cars in an infinitely repeati [#permalink]
21 Feb 2018, 07:14
Question Stats:
33% (01:18) correct
66% (01:44) wrong based on 12 sessions
A certain machine produces toy cars in an infinitely repeating cycle of blue, red, green, yellow and black. If 6 consecutively produced cars are selected at random, what is the probability that 2 of the cars selected are red? A. \(\frac{1}{6}\) B. \(\frac{1}{5}\) C. \(\frac{1}{3}\) D. \(\frac{2}{5}\) E. \(\frac{1}{2}\) We need to find possible and desired outcomes. What are the possible outcomes here? Well, we're going to be randomly selecting 6 consecutively produced cars and we want to know what the probability is that 2 of those cars will be red. How many different groups of 6 consecutive cars are there? Well, we've got 5 colors, so any one of them could be the first in the string of 6 that we choose. That gives us 5 different strings of 6. So 5 is the number of possible outcomes here. What about desired outcomes?
We need to know how many of those 5 strings have 2 red cars. What does getting two red cars depend on? It depends on the color of the first car of the string of 6. For example, if blue is the first car of the string, then you'd get blue, red, green, yellow, black and then blue again.
So, the only way to get 2 red cars is if the first car of the string is red, because then we'd have red, green, yellow, black, blue and then red again as the sixth car. So 1 out of the 5 strings will contain 2 red cars, so that's a probability of 1/5
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Re: A certain machine produces toy cars in an infinitely repeati [#permalink]
21 Feb 2018, 10:34
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This problem is a disguised version of the following question: if you have pick any 6 consecutive integers, what are the odds that two of them are divisible by 5? It's good to know that every fifth integer is divisible by 5, every 7th divisible by 7, etc. Why is this so? If we're trying to find the odds that two are divisible by 5, we could select some groups of 6 integers and see what happens: 123456 234567 345678 456789 5678910 Notice every set of integers has one 5 in it, but in the last set, if we start with a 5, we'll end on a 10, giving us two integers divisible by 5. If we scoot one over again we'll lose the 5 and have a 10 in its place for the next 4 sets. Long story short, every fifth group has two numbers divisible by 5. Let's apply it to this problem: BRGYBB RGYBBR GYBBRG YBBRGY BBRGYB Since these are the only 5 ways you can select 6 cars in this order, and since only one has a red car twice, the answer is 1/5, or B.
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Re: A certain machine produces toy cars in an infinitely repeati [#permalink]
19 Feb 2019, 06:17
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I found like this,
Cycle 1. blue, red, green, yellow and black Cycle 2. blue, red, green, yellow and black Cycle 3. blue, red, green, yellow and black and on and on....
So, every fifth car will be red.
blue, red, green, yellow and black are 5 cars. For selection for 6 cars set we need to consider two cycles at least.
Example: Blue to Blue will have 6 cars. Red to Red will have 6 cars and on and on....
blue, red, green, yellow, black, blue, red, green, yellow, black....
Now, we have 5 different cars getting produced in repeating cycle. Any time we pick 6 cars, for sure two cars with same color will be picked. As we have 5 different colors, chances for red to be picked is \(\frac{1}{5}\).
Ans. B




Re: A certain machine produces toy cars in an infinitely repeati
[#permalink]
19 Feb 2019, 06:17





