bowler has taken 5 five-wicket hauls in his last 17 matches. His match records are selected at random, one by one, and analyzed. If none of the match records is analyzed more than once, then what is the probability that the 11th one analyzed is his last five-wicket haul match? term ---a five wicket haul is a match in which a player took 5 wickets
so i am looking for way to calculate the probability using just combinations,i.e. is not including the notion of conditional probability
I've come up with (12C6 x 5C4)/17C10
but the answer is (12C6 x 5C4)/(17C10*7)
For simplicity, I will refer to the "five-wicket hauls" as outcomes labeled $A$ and the remaining games as outcomes labeled $B$
There are $\binom{17}{5}$ ways to pick the order of $A$'s and $B$'s such that there are five $A$'s and twelve $B$'s. Each of these orders of $A$'s and $B$'s are equally likely to have occurred when ordering and analyzing the match records.
Of these, $\binom{10}{4}$ of these are ways in which four of the $A$'s occur within the first ten spots and the fifth occurring on the $11$'th position.
The probability is then:
$$\frac{\binom{10}{4}}{\binom{17}{5}}=\frac{15}{442}\approx 0.03393665\dots$$
Now, comparing to yours and the book's answer... we see they both treated each match still as distinct without making the same simplification that I had. The numerators both suggest that we are trying to piece together what the first ten matches looked like. Both correctly count this as $\binom{12}{6}\binom{5}{4}$ ways to have picked six "B" type games and four "A" games (but keeping their specific match details distinct).
That is all well and good talking about the first ten matches, but we are talking about the first eleven matches. The eleventh match is going to be the remaining "$A$" game. Now, the denominator... in your answer was just talking about the number of ways the first ten matches could have been collectively selected.
Again... we are supposed to have been talking about the number of ways the first eleven matches were selected with special emphasis on which of those eleven selected happened to be the eleventh match.
The book's answer does this correctly by first picking the first ten matches in $\binom{17}{10}$ ways and then picking the eleventh match out of the seven remaining matches available in $7$ ways for a total of $\binom{17}{10}\cdot 7$. You could also have done this by picking the eleven matches and picking one from that in $\binom{17}{11}\cdot 11$ or by picking the eleventh match first and then picking the ten after in $17\cdot \binom{16}{10}$... there are several equally correct ways to do this... but in each we still remembered to keep track of the eleventh match somehow which your answer failed to do.