Mechan has a jar containing 15 counters. There are only blue counters, green counters, and red counters in the jar. Metor is going to take at random one of the counters from his bag of 12 counters. He will look at the counter and put it back into the bag. Meghan will then take at random one of the counters from her jar of counters.

a. What is the probability that the 3 counters each have a different color?

b. How many blue counters are there in the jar?

a. The probability that the 3 counters each have a different color is: 1/2 * 7/12 * 2/5 = 7/100b. Work out how many blue counters there are in the jar: There are a total of 3 colors. Therefore, it is given that: Blue + Red + Green = 15Let the number of blue counters be B. Therefore, the number of red counters = R and the number of green counters = G. Thus, B + R + G = 15

(i)Now, the probability that the 3 counters each have a different color is given as follows: P(BRG) = P(B) * P(R) * P(G/B and R)There are 3 ways in which we can have a jar with different colored balls: Blue, Red and Green; Red, Blue and Green; Green, Red and Blue. In each of these cases, there will be the same probability that each one of these cases would occur. Hence we need to multiply the probability of one of them by 3.Below is the probability distribution of selecting 1 counter from the bag at random: Blue = 3/12 = 1/4Red = 4/12 = 1/3Green = 5/12

Let us consider the case of selecting 1 counter from the bag at random with all colors having a different number of counters. There is a 1/4 chance of selecting a blue counter. Once a blue counter has been chosen, there will be 2 blue counters left in the jar. Hence, there will be 11 counters left in the jar of which 4 will be red. There is a 4/11 chance of selecting a red counter. Once a red counter has been chosen, there will be 3 red counters left in the jar. Hence, there will be 10 counters left in the jar of which 5 will be green.

There is a 5/10 chance of selecting a green counter. The probability that the 3 counters each have a different color is: P(BRG) = 1/4 * 4/11 * 1/2 = 1/22The probability that any 2 colors will be present will be the sum of the probability that BRG and that the probability that RGB will be drawn. P(BRG, BRG) = 1/22P(BRG, RGB) = 1/22P(RBG, RGB) = 1/22The probability that any 2 of the three colors will be present = 1/22 + 1/22 + 1/22 = 3/22

The probability that all 3 counters will have the same color is: P(BBB) = 3/12 * 2/11 * 1/10 = 1/220P(GGG) = 5/12 * 4/11 * 3/10 = 6/220P(RRR) = 4/12 * 3/11 * 2/10 = 1/55The total probability of getting the same color = 1/220 + 6/220 + 1/55 = 1/20The probability that the 3 counters each have a different color is: 1/22The probability that any 2 of the three colors will be present = 3/22 The probability that the 3 counters each have the same color is: 1/20 Given that there are 15 counters in the jar, then: B + R + G = 15

(i)Also, it is given that there are 12 counters in the bag. Therefore: B + R + G = 12We can subtract equation (i) from equation (ii) to obtain:0B + 0R + 0G = -3Thus, the equation is inconsistent and there are no solutions. Therefore, there are no blue counters in the jar.

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