Three cards are drawn successively,without replacement from the pack of 52 cards.Then the probability that the drawn card are face card of same suits , is ?
My approach : Probability of select one suit among 4 is 1/4.
Among each suits, there are 3 face card so the probability of face cards among each suits is (3/13)(2/12)(1/11).
So Required probability is 1/4 * (3/13)(2/11)(1/11).
Is the above approach correct?
Please help me.
First, one thing to note is that the question indicates no replacement this is usually a good hint for you to think about the hypergeometric distribution as that is the only probability distribution with no replacement.
In general, the distribution takes the following form:
$\frac{{K \choose k} {N-K \choose n-k}}{N \choose n} $
Where:
$N$ is the population size,
$K$ is the number of success states in the population,
$n$ is the number of draws (i.e. quantity drawn in each trial),
$k$ is the number of observed successes
The probability of drawing $3$ face cards of the same suit is then the following:
$\frac{{4 \choose 3} {48 \choose 0}}{52 \choose 3} =\frac{{4 \choose 3} }{52 \choose 3}$
Note that this counts the ace as a face card, if you don't want to count it that would be
$\frac{{3 \choose 3} {48 \choose 0}}{52 \choose 3} =\frac{{3 \choose 3} }{52 \choose 3}$
Then this should be multiplied by $4$ since that is the probability of drawing all $4$ face cards from one specific suit (so for example that is the probability of drawing all heart face cards, we multiply by $4$ to take into account ace, spades, and I forgot what the last suit is called.
so our final answer would be $4 \cdot \frac{{4 \choose 3} }{52 \choose 3}$ if we count the ace and $4 \cdot \frac{{3 \choose 3} }{52 \choose 3}$
Note that this isn't the only way to do it, the other way would be using multiplication rule, we can do the following (here I am not counting ace as a face card:
$\frac{3}{52}\cdot \frac{2}{51}\cdot\frac{1}{50}$ that is, probability of getting the first face card from a specific deck multiplied by the probability of getting the second face card from the same deck multiplied by the probability of getting the third face card from the same deck and then multiply the final answer by $4$ again (due to the same explanation) which would give the same result, note that since you are drawing these cards with no replacement the denominator is decreasing.
