On the school debate team, $6$ out of the $14$ girls are seniors and $9$ of the $20$ boys are seniors. What is the probability that either a girl or a senior would randomly be selected to argue a position?

My attempt
Let $G$ be the number of girls and $S$ the number of seniors. Then $$P(G \cup S) = P(G) + P(S) - P(G \cap S)$$ I don't know if I did this right or not.

So $$P(G \cup S) = (14/34) + (1/15) - (6/14)$$ Because there are $34$ total students and the chances of it being girls are $14$ over $35$, the chances of it being senior are $1$ over $15$ since $9 + 6 = 15$ and the chance of being senior is $1/15$.
I think I got $6/14$ because $P(S \cap G)$ would be $6/14$ because that's intersecting right? Did I do this right? How would I go on to solve this?

  • $\begingroup$ $1/15$ is not the probability that a randomly selected student is a senior; it is the probability that, given that one is a senior and a senior was selected, one was that selected senior. $6/14$ is not the (joint) probability that a senior girl was selected; it is the (conditional) probability that, given that a girl was selected, that girl was a senior. $\endgroup$ – Brian Tung Dec 17 '15 at 0:02
  • $\begingroup$ In addition to the suggestions and corrections given on the page, you can also solve it by finding the probability of selecting a nonsenior boy and subtracting that from one. $\endgroup$ – turkeyhundt Dec 17 '15 at 0:04

Assuming that each of the $34$ students is equally likely to be selected, the only way that the proposition fails to be satisfied is if a non-senior boy is selected. That happens $20-9 = 11$ times out of $34$. Therefore, the probability that the proposition is satisfied is $23/34$.

ETA: The general form of your approach is fine, but you have erred in your assignment of probability values for $P(S)$ and $P(G \cap S)$. Note that $G \cap S \subseteq S$ structurally, so $P(G \cap S) \leq P(S)$ necessarily, which is violated by your assignment.

In fact, the actual probability that the selected student is a senior is $(6+9)/(14+20) = 15/34$, and the probability that the selected student is a senior girl is $6/34$. Then your expression yields the correct value of

$$ P(G \cup S) = \frac{14}{34}+\frac{15}{34}-\frac{6}{34} = \frac{23}{34} $$

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  • $\begingroup$ Thanks a lot for your answer, I am kind of confused on $$ P ( A ∩ B )$$ Would you be able to explain it in a easier way and also i don't know what $$ ⊆$$ The symbol means ? $\endgroup$ – MATH ASKER Dec 17 '15 at 0:04
  • $\begingroup$ The symbol $\subseteq$ means "is a subset of"; in this case, it refers to the fact that the set of girls who are seniors is a subset of the set of seniors. I don't know what confusion you are having on $P(A \cap B)$, since there is no $A$ and $B$ defined anywhere. $\endgroup$ – Brian Tung Dec 17 '15 at 0:52
  • $\begingroup$ No, it was a general question, I am kind of confused in finding $$ P(A n B) $$ $\endgroup$ – MATH ASKER Dec 17 '15 at 1:01
  • $\begingroup$ It depends. In some cases, as here, you can simply tally up the set $A \cap B$, and divide by the number of total possibilities. In other cases, you may take advantage of an independence between $A$ and $B$, and obtain $P(A \cap B) = P(A)P(B)$. There is no sure-fire formula that covers all the possibilities; one is best just apprehending what the expression represents, and determining the best way to go about obtaining the value. $\endgroup$ – Brian Tung Dec 17 '15 at 2:14

How did you find the probability $P(S)$ again? There are $15$ seniors out of $34$ students. The probability $1/15$ would be the chance of picking one particular senior out of all the seniors, but you want the probability of picking any senior out of everybody.

Similarly, when calculating $P(S\cap G)$, you want the $6$ girls who are seniors out of everybody, not just the girls.

Also, I think you flipped the minus sign to a plus sign by accident in your calculation.

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