Conditional Probability (Different Context)

Question

A male customer visiting the suit department of a certain store will purchase a suit with probability 0.16, a shirt with probability 0.18, and a tie with probability 0.2. The customer will purchase both a suit and a shirt with probability 0.07, both a shirt and a tie with probability 0.09, and with no probability of purchasing a suit and a tie.

(a) What is the probability that a customer purchases at least one of these items?

P = 0.16 + 0.18 + 0.2 - 0.09 - 0.07 = 0.38

(b) Given that the customer purchases a shirt, what is the probability that he also purchases a tie?

P = 0.09/0.18 = 0.5

(c) Given that the customer purchases at least one item, what is the probability that he purchases a shirt and a tie?

I am confused about this part on conditional probability, as the context is different from part b. Would be very grateful to receive any constructive feedback regarding my question. Thank you!

Answer 1

(c) would be $\frac{0.09}{0.38}$ based on the probability under (a). So about $0.234$ or exactly $\frac{9}{38}$ (which cannot be simplified).

Or look at it this way: we can draw a Venn diagram (for 100 a costumers, say) for three sets Suits, Shirts and Ties, with thriple intersection having 0, Suit $\cap$ Shirts having $7$ elements, Suit $\cap$ Ties having $0$ and Shirts $\cap$ Ties having $9$ elements. The "unique" part of Shirts has $2$ elements ($18 - 9 - 7$), the unique part of Suits has $9$ ($16-7$) and for Ties $11$ ($20 - 9$). So all disjoint parts together have $9+7+2+9+11 = 38$ members buying something, nicely corroborating $a)$. There are $100 - 38 = 62$ customers outside the three sets. As to $b)$, Shirts has $18$ members, $9$ of which also are in Tie, so we get $\frac{1}{2}$.

Now of the $38$ buying at least something, $9$ buy a shirt and a tie. So given that we pick a customer from inside the Venn diagram, we clearly get $\frac{9}{38}$ as the answer.