Find the value of p (Probability)

[TestKing1994]

100 crates of oranges imported from Country A are inspected. The number of rotten oranges in the crates is recorded.



d) A crate is selected at random from the crates of oranges imported from Country A. The probability that the crate contains no fewer than p rotten oranges is 3/4. Find the value of p.

 

[ksdhart2]

That sure is a nice homework problem you've got there - but what are your thoughts? What have you tried? Please re-read the Read Before Posting thread that's stickied at the top of each sub-forum and comply with the rules therein. Specifically, please share with us any and all efforts you've made on this problem, even the parts you know for sure are wrong. Thank you.

Additionally, it will be very helpful if you can also include the problem text, and your workings, for the other three parts of this problem. I often find that the parts of a problem are intimately related, such that the work you've done on one part directly helps you figure out what to do on the next.

 

[TestKing1994]

100 crates of oranges imported from Country A are inspected. The number of rotten oranges in the crates is recorded.

View attachment 12387

c) A crate is selected at random from the crates of oranges imported from Country A. The probability that the crate contains no fewer than p rotten oranges is 3/4. Find the value of p.

100 crates of oranges imported from Country A are inspected. The number of rotten oranges in the crates is recorded.

View attachment 12387

d) A crate is selected at random from the crates of oranges imported from Country A. The probability that the crate contains no fewer than p rotten oranges is 3/4. Find the value of p.

Below is the added two parts for this question. The numbering is wrong though, it is supposed to be a,b, and c.

a) State the largest number of rotten oranges found in a crate from Country A.
Ans: Country A = 9 rotten oranges

b) Find the total number of rotten oranges from Country A.
Ans: Country A = the summation of rotten oranges of 100 crates
= (0*4) + (1*9) + (2*12) + (3*28)........(9*1)
= 349 rotten oranges

 

[TestKing1994]

That sure is a nice homework problem you've got there - but what are your thoughts? What have you tried? Please re-read the Read Before Posting thread that's stickied at the top of each sub-forum and comply with the rules therein. Specifically, please share with us any and all efforts you've made on this problem, even the parts you know for sure are wrong. Thank you.

Additionally, it will be very helpful if you can also include the problem text, and your workings, for the other three parts of this problem. I often find that the parts of a problem are intimately related, such that the work you've done on one part directly helps you figure out what to do on the next.

I've added the rest preceding parts with answers for this question. But I've no idea for part c which I've written here as d. My guess is that the probability for the existing crates with rotten oranges should be left with 96/100 crates since 4 crates do not contain rotten oranges. I know that the symbol for probability with p value should be written as P( no. rotten oranges > p).....But I have no idea for the proceeding parts.

 

[Dr.Peterson]

I would add a third row to the table, representing the cumulative number of crates -- that is, the first column will be 4, meaning that 4 crates had no more than 0 rotten; the second will be 4+9 = 13, meaning that 13 had no more than 1 rotten; and so on.

Then a fourth row could be the cumulative probabilities: column p is the probability of having no more than p rotten.

Then you can figure out the value of p for which this probability is ... what? Or you might prefer just to decide what number in the third row matches the probability you are looking for (to save a little calculation).

Alternatively, you might work out that third row from right to left, which better matches what you are looking for: the probability that the number of rotten oranges is greater than p.