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!

  • $\begingroup$ What is the probability that the customer buys nothing at all? $\endgroup$
    – user694818
    Aug 24, 2019 at 12:38
  • $\begingroup$ probability will be 1 - P(at least one item) = 1 - 0.38 = 0.62 $\endgroup$
    – kekeke12
    Aug 24, 2019 at 12:42

1 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.

  • $\begingroup$ sorry, but could you explain why using the probability 0.09 to solve the answer as the context for probability of purchasing both a shirt and tie is 0.09 when the customer enters the store and not in the context where the customer buys at least one item $\endgroup$
    – kekeke12
    Aug 24, 2019 at 12:50
  • $\begingroup$ that is to say there will be a 0.91 probability that the customer will not purchase both shirt and tie, while the same could not be said should the customer buy at least one item $\endgroup$
    – kekeke12
    Aug 24, 2019 at 12:53
  • $\begingroup$ @kekeke12 one event is a subset of the other $\endgroup$ Aug 24, 2019 at 12:53
  • $\begingroup$ @kekeke12 added an extra explanation. No idea how to draw a Venn diagram on this site, so you do the drawing. $\endgroup$ Aug 24, 2019 at 13:08

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