In a box,there are 8 alphabets cards with the letters S,S,A,A,A,H,H,H.Find the probability that the word ASH will form if the three cards are drawn simultaneously.

I can count the total no of cases$=\binom{8}{3}$ but i cannot count the favorable cases.The answer for the question is $\frac{9}{28}$


For the favorable cases, assuming (as you did for the counting of total cases) that the cards are distinguishable and that extraction order does not matter [*] :

You need one card of each letter. You have $2$ alternatives for a $S$ card, $3$ for each of the others. Then there are $2 \times 3\times 3 = 18 $ favorable cases.

Then $p=\frac{18}{\dbinom{8}{3}}=\dfrac{18}{56}=\dfrac{9}{28}$

[*] Note that you can change this assumption, which could alter the counting of favorable vs total cases... but, of course, not the probability. For example, counting the three extractions as ordered, we'd have a total of $8 \times 7 \times 6 = 336$ possible extractions; and the favorable ones would be $ (2 \times 3\times 3) 3!=108$ , where the first factor count the ways of getting the (ordered) sequence SAH, and the factorial takes into account the permutations. Then, again $p=\frac{108}{336}=\frac{9}{28}$.

  • $\begingroup$ You have a typo at the final result. $\endgroup$ – callculus Jul 19 '18 at 17:15

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