A certain machine produces toy cars in an infinitely repeating cycle

A. \(\frac{1}{6}\)
B. \(\frac{1}{5}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{2}\)

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....

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....

For each of the colors we have 5 sets with 6 cars each. Now for selecting 6 cars out of which 2 are Red can happen if out of these 5 sets we choose the set which has Red to Red in it. Shown below:

Cycle 1. blue, red, green, yellow and black Cycle 2. blue red, green, yellow and black

So, probability \(\frac{1}{5}\).

Other way to look at the problem is that 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 \(\frac{1}{5}\).

Answer: (B).

TOTAL Variants:1st cycle Blue/Red/Green/Yellow/Black

2nd cycle same order:Blue/Red/Green/Yellow/Black : total production :10

Pbt of getting red : 2/10 = 1/5 (B)

The cars are produced in infinitely repeating cycles of blue, red, green, yellow and black.
We are told that 6 consecutive cars are selected at random. We have been asked to find
the probability that there are 2 red cars. This is possible if the first car is red in color.

Since the car is available in 5 colors, the probability of the first car being red in color is \(\frac{1}{5}\)(Option B)

Solution

Given:
• A certain machine produces toy cars in an infinitely repeating cycle of blue (BU), red (R), green (G), yellow (Y), and black (BL)

To find:
• If 6 consecutively produced cars are selected at random, what is the probability that 2 of the cars selected are red

Approach and Working:
• As the color of the cars follow a specific sequence of 5 definite colors, which is getting repeated infinite times, the following are the possible sequences of the repeating cycles:

o BU, R, G, Y, BL, BU …
o R, G, Y, BL, BU, R …
o G, Y, BL, BU, R, G …
o Y, BL, BU, R, G, Y …
o BL, BU, R, G, Y, BL …

As we can see, these are the 5 possible sequences at which the colors are getting repeated
• Hence, total number of cases = 5

Now, if we observe the possible sequences, in every case the 6th car’s color is getting repeated as the color of the 1st car
Therefore, if 6 consecutively produced cars are selected at random, getting 2 cars selected as R is possible when the 1st color is also R (i.e. R G Y BL BU R …)
• Hence, favorable cases = 1
• Therefore, the required probability = \(\frac{1}{5}\)

Hence, the correct answer is option B.

Answer: B

The same colored car is repeated after every 5 cars i.e. every sixth car is the same color. Hence, probability of picking red first = 1/5
If red is picked first, the sixth car will anyway be red.

Hence, B.

P(2 reds) = ?

Re-phrased, the question is asking us what is the probability of picking a red on the first and last attempt:

R G Y Bl Blu R <--- This is the only possible way that we will get 2 reds given that 6 are chosen

Now, since there are 5 colors, the denominator is 5.

P(2 reds) = 1/5

Answer is B.

Imagine it like a question where 6 places need to be filled by available number of cars of 5 types.

Total scenarios: 1st space can be filled in 5 ways, 2nd place can be filled in 4 ways, 3rd place can be filled in 3 ways, 4th place can be filled in 2 ways, 5th place can be filled in 1 way.

The 6th place would automatically get filled in 1 way by the same coloured car which takes the initial space.

Thus, total ways = 5!X1 = 120

Now, there would be only 1 scenario of selecting two red cars. i.e When red colour car takes the 1st place.

So, the spaces can be filled as
1*4*3*2*1*1 = 24

Thus,
Probability= 24/120 = 1/5

Hope this helps.
Paritosh Rawat

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