Help with probability

[Jaskaran]

Hello!

I am confused on how to approach this "simple" yet mind-bogging probability problem.

You pick three cards at random from a deck. Find the probability of each event described below:
a) You get no Aces (there are 4 aces in a deck of 52..so the probability of getting now aces would be 48/52? )
b)You get all hearts (there are 13 hears in a deck of 52, so (13/52)*(12/51)*(11/50) is the probability?
c) The first card is your first red card (the probability of it not being a red card is 26/52, so the probability is(1 - (26/52)*(26/52))?
d) You have at least one diamond. So the probability of it not being a diamond is 36/52, so the probability is (36/52)*(36/52)*(13/52)?

Am I doing plenty wrong?

Second problem...

A private college report contains these statistics"

70% of incoming freshmen attended public schools.
75% of public school studetns who enroll as freshmen eventually graduate.
90% of other freshmen eventually graduate.

a) Is there any evidence that a freshman's chances to graduate may depend upon what kind of High School the student attended? Explain But how?

b)What perecent of freshmen eventually graduate.How can I approach this?

Thank you for all your input and help, thanks.

Jaskaran.

For the first problem, I get:
a).782
b).012
c).122
d).58

Am I right or wrong for the first problem? Help is appreciated.

 

[soroban]

Hello, Jaskaran!

You pick three cards at random from a deck.
Find the probability of each event described below:

a) You get no Aces
There are 4 aces in a deck of 52,
so the probability of getting no Aces would be 48/52? . . . no

That would be for picking one card.

For picking three non-Aces: \(\displaystyle \L\:\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\)

b)You get all hearts
There are 13 hearts in a deck of 52,
so (13/52)*(12/51)*(11/50) is the probability?

Right!

c) The first card is your first red card
The probability of it not being a red card is 26/52,
so the probability is(1 - (26/52)*(26/52) ? . . . no

The probability that the first card is red is \(\displaystyle \,\frac{26}{52}\,=\,\frac{1}{2}\)
. . and we don't care what the next two cards are, right?

The answer is: \(\displaystyle \L\:\frac{1}{2}\)

d) You have at least one diamond.
So the probability of it not being a diamond is 36/52,
so the probability is (36/52)*(36/52)*(13/52)? . . . no

"At least one diamond" means: one diamond, two diamonds, or three diamonds.
. . We must find these three probabilities and add them.

It is easier to find the "opposite": the probability of no diamonds.

There are 39 non-diamonds is in the deck.
The probability of picking three non-diamonds is: \(\displaystyle \L\:\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\)
Then subtract that result from \(\displaystyle 1.\)