Probability - 5 card hand

Question

Question is :

You have a 5 card hand from randomly shuffled standard deck of 52 cards. P - Event that hand exactly contains one spade.
Q - Event that hand exactly contains one ace.

Calculate :
a. Prob[P]
b. Prob[Q]
c. Prob[P intersection Q]

I've faced this kind of problem for the first time. Can anyone please guide me ?

Answer 1

Can you calculate the probability of this event

S N N N N

where S stands for spade & N for Not spade.

But this is just one case. Spade could appear at any point from 1 to 5.

So multiply the probability with 5. This will be answer to part (a)

Similar logic applies to part (b)

Answer 2

We use the Use Hypergeometric Distribution:

$P(One$ $Spade)$ = $\dfrac{\binom{13}{1}\binom{39}{4}}{\binom{52}{5}}$ $=$ $\dfrac{{13} \cdot {82,251}}{2,598,960}$ $= 0.41$.

$P(One$ $Ace)$ = $\dfrac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}$ $=$ $\dfrac{{4} \cdot {194,580}}{2,598,960}$ $= 0.30$.

Then modify Hypergeometric for intersection:

$P(One$ $Spade$ $\cap$ $One$ $Ace)$ = $\dfrac{\binom{4}{1}\binom{13}{1}\binom{39}{3}}{\binom{52}{5}}$ $=$ $\dfrac{{4} \cdot{13} \cdot {9,139}}{2,598,960}$ $= 0.18$.