# Thread: 2 probability questions...

1. ## 2 probability questions...

1. You are dealt 5 cards from a standard 52 card deck, what is the condition probability that the hand contains no pairs and no spades?
2. The probability of snow is 20%, and the probability of a traffic accident is 10%. Suppose further that the conditional probability of an accident, given that it snows, is 40%. What is the conditional probability that it snows, given that there is an accident?

2. 1) $\displaystyle \frac{(39)X(38-2)X(37-4)X(36-6)X(35-8)}{52X51X50X49X48}$

2) Is it 80% ?

3. Hello, Sneaky!

I agree with Traveller's answers.

1. You are dealt 5 cards from a standard 52-card deck, .What is
the probability that the hand contains no pairs and no Spades?

I had to baby-talk my way through this one . . .

We begin with a deck with 39 cards . . . only Hearts, Clubs, and Diamonds.

The 1st card can be any of the 39 cards: .$\displaystyle \boxed{39}\text{ choices}$
(Suppose it is the $\displaystyle 3\heartsuit$.)

We draw a 2nd card from the remaining 38 cards in the deck,
. . but we do not want the $\displaystyle 3\clubsuit\text{ or }3\diamondsuit:\;\boxed{36}\text{ choices}$
(Suppose it is the $\displaystyle 6\clubsuit$.)

We draw a 3rd card from the remaining 37 cards in the deck.
. . but we do not want: $\displaystyle 3\clubsuit,\:3\diamondsuit,\:6\heartsuit,\,6\diamo ndsuit :\;\boxed{33}\text{ choices}$
(Suppose it is the $\displaystyle 9\diamondsuit$,)

We draw a 4th card from the remaining 36 cards in the deck.
. . We do not want: $\displaystyle 3\clubsuit,\:3\diamondsuit,\:6\heartsuit,\:6\diamo ndsuit,\:9\heartsuit,\:9\clubsuit:\;\boxed{30}\tex t{ choices}$
(Suppose it is the $\displaystyle J\heartsuit$.)

We draw a 5th card from the remaining 35 cards in the deck.
. . We do not want: $\displaystyle 3\clubsuit,\:3\diamondsuit,\:6\heartsuit,\:6\diamo ndsuit,\:9\heartsuit,\:9\clubsuit,\:J\clubsuit,\:J \diamondsuit: \;\boxed{27}\text{ choices}$

Hence, there are: .$\displaystyle 39\cdot36\cdot33\cdot30\cdot27 \:=\:37,\!528,\!920$ hands
. . with no pairs and no Spades.

And there are: .$\displaystyle 52\cdot51\cdot50\cdot49\cdot48 \:=\:311,\!875,\!200$ possible hands.

$\displaystyle \text{Therefore: }\;P(\text{no pairs and no Spades}) \;=\;\dfrac{37,\!528,\!920}{311,\!875,\!200} \;=\;\dfrac{8,\!019}{66,\!640}$