I am currently doing a self-study on Conditional Probability. I was faced with a question where I was provided with $P(a)$, $P(b)$, $P(c)$, $P(a \mid d)$, $P(b \mid d)$ and $P(c \mid d)$.

The full context of the question:

A departmental store reports that 30% of payment is collected in cash, 60% in credit card and 10% in debit card. 20% of cash purchases, 90% of credit card purchases and 80% of debit card purchases are for more than 200 dollars in purchases. What the probability that Sue paid cash if she purchased a new bag that costs $98.

Thus, I took $a = \mbox{payment in cash}$, $b = \mbox{payment in credit}$ and $c = \mbox{payment as debit}$. $P$ represented the probability that payments went above 200 dollars.

I was attempting to find out what is $P(d)$.

My attempt (which did not seem to yield the right answer): $$ P(a)P(a \mid d) + P(b) P(b \mid d) + P(c)P(c \mid d) $$

Appreciate some guidance please.


Based on the latest guidance, I still failed to answer the question with the final answer being 1.43567, however still deferring from the original answer of 0.750. My answer seemed grossly wrong.

My solution:

Let $a = \mbox{cash}$, $b = \mbox{credit}$, $c = \mbox{debit}$ and $d = \mbox{payment amounts more than 200 dollars}$.

Therefore $P(a) = 0.3$, $P(b) = 0.6$, $P(c) = 0.1$, $P(d \mid a) = 0.2$, $P(d \mid b) = 0.9$ and $P(d \mid c) = 0.8$.

I will first attempt to find \begin{align*} P(d) &= P(a)P(d \mid a) + P(b)P(d \mid b) + P(c)P(d \mid c) \\ &= 0.3 \cdot 0.2 + 0.6 \cdot 0.9 + 0.1 \cdot 0.8 = 0.68. \end{align*}

Then, I will attempt to find the value of $P(a \mid d)$ using the bayes theorem which results in $(0.2+0.3)/0.68 = 0.73529$.

Next, to find $P(a \mid d^c)$, we use the formula $P(d) = P(a)P(a \mid d) + P(d^c) P(a \mid d^c)$. $0.68 = 0.3 \cdot 0.73529 + 0.32 P(a \mid d^c) = 1.43567$.

Since the probability can never be more than $1$, my answer looks terrible wrong. Appreciate and guidance please.

  • 1
    $\begingroup$ Start by drawing a diagram (have you included all relevant information in your question)? Why would you expect your attempt to yield P(D)? What facts did you use in constructing that answer? $\endgroup$
    – Glen_b
    Apr 27, 2014 at 6:32
  • $\begingroup$ Hi Glen. Thanks for responding. I'd edit the question abobe to add the context. $\endgroup$ Apr 27, 2014 at 7:56
  • 1
    $\begingroup$ Do you notice the essential difference between what you have in your comment and what you have in your question? $\endgroup$
    – Glen_b
    Apr 27, 2014 at 9:19
  • 1
    $\begingroup$ Are you really sure about the computation of $P(a \mid d)$? $\endgroup$
    – QuantIbex
    Apr 27, 2014 at 10:49
  • 1
    $\begingroup$ This question still needs to be fixed. Was the discussion of Bayes theorem of any use? $\endgroup$
    – Glen_b
    Apr 29, 2014 at 0:36

2 Answers 2


There are several ways to approach this question, but I believe the intent is to use Bayes' Theorem:

$$P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}$$

specifically the version where the denominator is split up:

$$P(A_i|B) = \frac{P(B|A_i)\,P(A_i)}{\sum\limits_j P(B|A_j)\,P(A_j)}\cdot$$

Bayes' theorem is just the thing for swapping the direction of conditioning around (A|B in terms of B|A).

Aside from the fact that you're dealing with $D^c$ when you information is in terms of $D$, it's a very straightforward application of the theorem. You need to add an extra little step of thought to sort that out in the expression on the RHS.


thanks for the advice and inputs everyone. Following's my working:

enter image description here


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