question:
If an orange contains a seed, it fails the test. 1/5 oranges have seeds. 10
randomly selected oranges are tested, if even one orange contains a seed,
the test is failed.
a) What is the probability of the test passing?
b) The inspector decides he will inspect until 2 failures are found. What is the probability that 5 inspections are needed in order to reach the target of 2 failed inspections?
I figured that this might be a binomial distribution problem where you go until the first success, or in this case the first fail. I wasn't too sure how to structure this question. I thought about doing $$p = 1/5, x = 0, n = 10 $$
$$using \space p^x(1-p)^{n-x} $$
$$C_{10,0}(\frac{1}{5})^{0}(1-\frac{1}{5})^{10} = .1074$$
My reasoning is that 1/5 chance to fail, and $0$ fail in $10$ tests.
Can anyone verify if this is correct or i'm on the right path? Really confused whether or not to use geometric distribution or binomial. I think i'm supposed to use geometric for the second?
PART B ATTEMPT)
So using part a) $(1-.1074) = .8926$ (the probability in which a test will fail) So using Binomial $$C_{5,2}(.8926)^{2}(1-.8126)^{3}$$
What i wonder about the above formula is that does that represent 2 failures happening in 5 tests? or 2 failures happening BY the 5th test? or does it even matter?
Best Answer
Your answer is correct. The probability of the test passing is the probability that all oranges will pass the test, and since the oranges are independent, it is equal to the probability that the first orange will pass the test times the probability that the second orange will pass the test times etc,
meaning it is equal to $$(1-\frac15)^{10}\approx 0.1074$$
To answer (b), however, you will need more work. The probability of $5$ inspections being required to pass means you need to calculate the probability of the first four inspections failing and the last one succeeding. Using binomial distributions is the way to go, though, because as you may remember, a binomial distribution models the number of successes after $n$ independent binary tests, which is exactly what you have.