Question

The torque required to remove bolts in a steel plate is rated as very high, high, average, and low, and these occur about 30%, 40%, 20%, and 10% of the time, respectively. Suppose n = 25 bolts are rated; what is the probability of rating 7 very high, 8 high, 6 average, and 4 low? Assume independence of the 25 trials.

Answer 1

Answer:

0.004

Step-by-step explanation:

If two events A, B with probabilities P(A), P(B) are independent, then the probability of the two events ocurring at the same time A∩B is P(A)P(B).

In this case we have

P(torque=very high) = 0.3

P(torque=high) = 0.4

P(torque=average) = 0.2

P(torque=low) = 0.01

Using the product rule, we see there are

$\frac{25!}{7!8!6!4!}=4.4172*10^{12}$

ways of partitioning 25 bolts in groups of 7,8,6 and 4 bolts.

Each one of those partitions has a probability of

$(P(torque=very high))^7*(P(torque=high))^8*(P(torque=average))^6*(P(torque=low))^4= (0.3^7)*(0.4^8)*(0.2^6)*(0.1^4)=9.1729*10^{-16}$

So, the probability of rating 7 very high, 8 high, 6 average, and 4 low is

$4.4172*10^{12}*9.1729*10^{-16}=0.004$