Question

13. Recall the following situation, from a previous exercise: An assembly line worker’s job is to install a particular part in a device, a task which they can do with a probability of success of 0.78 on each attempt. (Assume that a success on one attempt is independent of success on all other previous or future attempts.) Suppose that they need to install ten such parts a day.

(a) What is the probability that it will take them 12 or more tries to install ten such parts? What you provide should be a slight modification of what you provide.

(b) What is the probability that, during a five day work week, it will take the worker 12 or more tries to install all the parts on exactly three of the days?You will need to use your result from part (a), along with a different distribution.

(c) What is the probability that, during a five day work week, it will take the worker less than 12 tries to install all the parts on one or two of the days?

Answer 1

Answer:-

Given That:-

Recall the following situation, from a previous exercise: An assembly line worker’s job is to install a particular part in a device, a task which they can do with a probability of success of 0.78 on each attempt. (Assume that a success on one attempt is independent of success on all other previous or future attempts.) Suppose that they need to install ten such parts a day.

Probability of success is 0.78 on each attempt.

(a) What is the probability that it will take them 12 or more tries to install ten such parts? What you provide should be a slight modification of what you provide.?

The given problem can be approached using Binomial distribution (however ,in this problem X is not Binomially distribution )as follows .

Suppose .random variable X denotes the number of attempts required for 10 successes,

To get 10 successes at least 10 attempts is necessary and hence

So, required probability is given by

The case X = 10 denotes the situation when all of the 10 attempts were successful.

The case X = 11 denotes the situation when one attempts of 1st to 10th was unsuccessful (i.e 9 attempts out of first 10 attempts were successful) and 11th attempts was successful .

(b) What is the probability that, during a five day work week, it will take the worker 12 or more tries to install all the parts on exactly three of the days?You will need to use your result from part (a), along with a different distribution.?

we can formulate the given problem as Binomial distribution as follows .

we consider taking 12 or more tries in a day as success.

Probability of taking 12 or more tries is 0.7332551.

Suppose,random variable Y denotes number of days in which 12 or more tries were needed out 5 days .

So,required probability is given by

(c) What is the probability that, during a five day work week, it will take the worker less than 12 tries to install all the parts on one or two of the days?

We can formulate the given problem as Binomial distribution as follows.

We consider taking less than 12 tries in a day as success.

Since at least 10 tries is compulsory, probability of taking less than 12 tries

Suppose, random variable Z denotes number of days in which less than 12 tries were needed out of 5 days.

So, required probability is given by​​​​​​​

Note- Part (c) also could be done alternatively as probability of 3 or 4 days with 12 or more tries to get 10 successes. In that case we have to calculate P(Y=3)+P(Y=4) and obviously that is same as our calculated probability.