# Probability for second sample - random selection without replacement

I am reproducing a probability question verbatim. I am unable to correctly solve the problem.

Question:

A batch of 25 injection-molded parts contains 5 parts that have suffered excessive shrinkage.

(a) If two parts are selected at random, and without replacement, what is the probability that the second part selected is one with excessive shrinkage?

(b) If three parts are selected at random, and without replacement, what is the probability that the third part selected is one with excessive shrinkage?

a). 5C2/25C2 + 20C1/25C1 * 5C1/25C1 = 29/150

b). Didn't try to solve as my answer to part (a) is wrong

a) 0.2

b) 0.2

What are different ways to approach solving this kind of problem ?

• You have committed a small error. In your solution to part (a), the second term on the LHS should be $\frac{\binom{20}{1}}{\binom{25}{1}}\frac{\binom{5}{1}}{\binom{24}{1}}$, since after picking one out of the $20$ without excessive shrinkage, there will be $24$ left. With this, the answer evaluates to $0.2$. – PN Karthik Aug 3 '16 at 7:22
• @Karthik: Thanks for pointing out the mistake. Now I am able to solve part b similarly. You may right the answer and I will upvote and select that as answer. BTW, can I solve this problem using conditional probability formulaes ? – nurabha Aug 3 '16 at 7:32
• I am happy that you were able to solve the problem :) I will not write it as the answer since you have been able to obtain the solution yourself. In what manner do you want to solve the problem using conditional probabilities? Can you show me the initial steps that you have worked out, if any? – PN Karthik Aug 3 '16 at 7:45
• A = Event that 1st part out of selected sample has excessive shrinkage, B = Event that 2nd part out of selected sample has excessive shrinkage, C = Event that 3rd part out of selected sample has excessive shrinkage. Initial steps for part a: P(B) = P(B AND A) + P(B and not A) = P(A)* P(B|A) + P(not A) * P(B|not A). – nurabha Aug 3 '16 at 8:12
• Yes, you can do that too. In some sense, that is precisely what you have done in your solution. For, $P(A)\cdot P(B|A)=\frac{\binom{5}{1}}{\binom{25}{1}}\frac{\binom{4}{1}}{\binom{24}{1}}$. The second term on the LHS of your solution to part (a) is precisely $P(A^{c})P(B|A^{c})$. – PN Karthik Aug 3 '16 at 9:20

I will try to reach unto you a new way of thinking.

Suppose that all parts are given randomly a distinct number from $\{1,2,\cdots,25\}$.

What is the probability that e.g. part numbered $17$ is one with excessive shrinkage?

It is $\frac5{25}=\frac15=0.2$ and this also for parts with other numbers. Every number has the same probability on this and $5$ of the $25$ parts are "gifted".

In (a) and (b) you are somehow giving numbers too. However in (a) you stop this if after giving $2$ numbers (the first selected gets number $1$, the second gets number $2$...) and in (b) after giving $3$ numbers. Nothing stops you from going on with the numbering until you reach $25$.

You can also think of $25$ numbered parts. Then $5$ of the numbers are randomly chosen and get the stamp "with excessive shrinkage". What is the probability then number $17$ is among those $5$? Same answer.

Take care of it that you master this way of thinking. It will bring you great profit in the study of probabilities.