2.The daily production from a car factory=3 different models-A,B,C which are in two different colours-R and S.
Daily the factory produces 1600 model A cars,800 of model B and 600 of model C. Overall,40% of the cars are of colour R,60% of colour S.60% of of the model A cars are of colour R.
As a percentage of production,there are 3 times as many colour R cars of model B compared to that of model C.
(a)Show that 192 model B cars of colour R are produced daily.
(c)If a car is chosen at random,calculate the probability that it is a model C car in colour S.
(d)Two cars are selected at random from a day's production.Calculate the probability that:
(i)neither of the cars is a model C in colour S.
(ii)at least one car's a model C in colour S.
(iii)the two cars are both model C but of different colours.
(iv)the two cars have the same colour.
Based the information given there are:
960 A-R
640 A-S
192 B-R
608 B-S
48 C-R
552 C-S
so,
c) = 552/3000
di) = (1.-552/3000) * (1.-551/2999)
dii) = 1.-(di)
diii) = (552/3000) * (48/2999)
div) = 1. - ( (1200/3000)*(1800/2999))
To solve this problem, we'll need to use the information given and calculate the required probabilities.
(a) First, let's calculate the number of model B cars of color R produced daily. We know that there are 1600 model A cars produced daily, and 60% of them are color R. So, the number of model A cars of color R is 0.6 * 1600 = 960.
It is given that there are 3 times as many color R cars of model B compared to that of model C. Let's assume the number of model C cars of color R is x. Therefore, the number of model B cars of color R would be 3x.
The total number of color R cars produced daily is the sum of model A, model B, and model C cars of color R, which is 960 + x + 3x = 960 + 4x.
We have been given that 40% of the cars produced are color R, which is 0.4 times the total production of each model:
0.4 * (1600 + 800 + 600) = 0.4 * 3000 = 1200.
Therefore, we can set up the equation:
960 + 4x = 1200.
Solving this equation, we find:
4x = 1200 - 960,
4x = 240,
x = 240 / 4,
x = 60.
So, the number of model B cars of color R produced daily is 3 times x, which is 3 * 60 = 180.
Therefore, 192 model B cars of color R are produced daily.
(c) To calculate the probability that a randomly chosen car is a model C car in color S, we need to find the number of model C cars of color S produced daily. We know the total production of model C cars is 600, and we also know that 60% of the cars are color S. Therefore, the number of model C cars of color S is 0.6 * 600 = 360.
The probability that a randomly chosen car is a model C car in color S is the ratio of the number of model C cars of color S to the total number of cars produced daily, which is:
360 / (1600 + 800 + 600) = 360 / 3000 = 0.12, or 12%.
Therefore, the probability that a randomly chosen car is a model C car in color S is 12%.
(d) Now, let's calculate the probabilities for the given scenarios.
(i) Probability that neither of the cars is a model C in color S:
To calculate this probability, we need to find the number of cars that are not model C cars in color S. We already know the number of model C cars of color S is 360. So, the number of cars that are not model C in color S is 3000 - 360 = 2640.
The probability that the first car chosen is not a model C car in color S is 2640 / 3000. After the first car is chosen, there are 2999 cars left, out of which 2639 are not model C cars in color S. Therefore, the probability that the second car chosen is also not a model C car in color S is 2639 / 2999.
To find the probability that neither of the cars is a model C in color S, we multiply the individual probabilities:
(2640 / 3000) * (2639 / 2999) = 0.9235.
Therefore, the probability that neither of the cars is a model C in color S is approximately 0.9235.
(ii) Probability that at least one car is a model C in color S:
To calculate this probability, we subtract the probability that neither of the cars is a model C in color S from 1.
Probability that at least one car is a model C in color S = 1 - Probability that neither of the cars is a model C in color S.
Probability that at least one car is a model C in color S = 1 - 0.9235 = 0.0765.
Therefore, the probability that at least one car is a model C in color S is approximately 0.0765.
(iii) Probability that the two cars are both model C but of different colors:
To calculate this probability, we need to find the number of ways to select two model C cars of different colors. There are two possible color options for the first car and only one remaining option for the second car. So, the number of ways to select two model C cars of different colors is 2 * 1 = 2.
The total number of ways to select two cars from a day's production is the combination of all the cars:
Total number of ways to select two cars = (1600 + 800 + 600) * (1600 + 800 + 600 - 1) = 3000 * 2999.
Therefore, the probability that the two cars are both model C but of different colors is:
2 / (3000 * 2999) = 1 / (1500 * 2999) ≈ 6.67e-4.
(iv) Probability that the two cars have the same color:
To calculate this probability, we need to find the number of ways to select two cars of the same color. There are two possible color options (R and S), so the number of ways to select two cars of the same color is:
Number of ways to select two cars of the same color = (1600 + 800 + 600) * (1600 + 799 + 599) + (800 + 600) * (799 + 599) + (600) * (599).
The total number of ways to select two cars from a day's production is the same as above, which is (3000 * 2999).
Therefore, the probability that the two cars have the same color is:
(Number of ways to select two cars of the same color) / (Total number of ways to select two cars) = [(1600 + 800 + 600) * (1600 + 799 + 599) + (800 + 600) * (799 + 599) + (600) * (599)] / (3000 * 2999).