The next generation of miniaturized wireless capsules with active locomotion will require
two miniature electric motors to operate each capsule. Suppose 20 motors have been invented
but that, in spite of tests performed on the individual motors, 4 will not operate satisfactorily
when placed into a capsule. If the scientist wishes to construct two capsules, with two motors
each, find the probability that among the four randomly selected motors
a) all four operate satisfactorily;
b) three operate satisfactorily and one does not.
Refer to Example 12 of motors for miniaturized capsules,
but instead suppose that 20 motors are available
and that 4 will not operate satisfactorily, when
placed in a capsule. If the scientist wishes to fabricate
two capsules, with two motors each, find the
probability that among the four randomly selected
motors
(a) all four operate satisfactorily;
(b) three operate satisfactorily and one does not.
To solve this problem, we will use the concept of combinations and probability.
a) Probability that all four motors operate satisfactorily:
There are 20 motors in total, and 4 of them do not operate satisfactorily. So, there are 16 motors that do operate satisfactorily. We want to find the probability that all four randomly selected motors from these 16 will operate satisfactorily.
The total number of ways to select 4 motors from the 16 satisfactory ones is given by the combination formula: C(16, 4) = 16! / (4! * (16-4)!). This calculates to be 1820.
The total number of ways to select 4 motors from the 20 available motors is given by: C(20, 4) = 20! / (4! * (20-4)!). This calculates to be 4845.
Therefore, the probability that all four randomly selected motors will operate satisfactorily is calculated as: 1820 / 4845 ≈ 0.376.
b) Probability that three motors operate satisfactorily and one does not:
Similar to part (a), we will first calculate the total number of ways to select 3 motors that operate satisfactorily from the 16 available motors, and 1 motor that does not operate satisfactorily from the 4 problematic motors.
The total number of ways to choose 3 satisfactory motors from 16 is given by: C(16, 3) = 16! / (3! * (16-3)!). This calculates to be 560.
The total number of ways to choose 1 problematic motor from 4 is given by: C(4, 1) = 4! / (1! * (4-1)!). This calculates to be 4.
Therefore, the total number of ways to select 3 satisfactory motors and 1 problematic motor is given by: (560 * 4) = 2240.
The total number of ways to select 4 motors from the 20 available motors is still C(20, 4) = 4845.
Therefore, the probability that three motors will operate satisfactorily and one will not is calculated as: 2240 / 4845 ≈ 0.462.
So, the probability for Part (a) is approximately 0.376 and for Part (b) is approximately 0.462.
To find the probabilities, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes.
a) To find the probability that all four randomly selected motors operate satisfactorily, we need to select 4 motors from the 16 motors that work.
Number of favorable outcomes: We need to select all four good motors, which can be done in 16 choose 4 ways.
Possible outcomes: We need to select four motors from a total of 20 motors, which can be done in 20 choose 4 ways.
The probability that all four motors operate satisfactorily is:
P(all four operate satisfactorily) = (16 choose 4) / (20 choose 4)
b) To find the probability that three motors operate satisfactorily and one does not, we need to choose 3 good motors and 1 bad motor.
Number of favorable outcomes: We need to select three good motors from the 16 that work and one bad motor from the 4 that do not work. This can be done in (16 choose 3) * (4 choose 1) ways.
Possible outcomes: We still need to select four motors from a total of 20 motors, which can be done in 20 choose 4 ways.
The probability that three motors operate satisfactorily and one does not is:
P(three operate satisfactorily and one does not) = [(16 choose 3) * (4 choose 1)] / (20 choose 4)