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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.

Edit

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.

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    $\begingroup$ 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
    Commented Apr 27, 2014 at 6:32
  • $\begingroup$ Hi Glen. Thanks for responding. I'd edit the question abobe to add the context. $\endgroup$
    – user1275515
    Commented Apr 27, 2014 at 7:56
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    $\begingroup$ Do you notice the essential difference between what you have in your comment and what you have in your question? $\endgroup$
    – Glen_b
    Commented Apr 27, 2014 at 9:19
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    $\begingroup$ Are you really sure about the computation of $P(a \mid d)$? $\endgroup$
    – QuantIbex
    Commented Apr 27, 2014 at 10:49
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    $\begingroup$ This question still needs to be fixed. Was the discussion of Bayes theorem of any use? $\endgroup$
    – Glen_b
    Commented Apr 29, 2014 at 0:36

2 Answers 2

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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.

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

enter image description here

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