Questions about Discrete Probability

[proff.hacker]

hi everybody,
I have 4 questions about Discrete Probability:

1- What is the probability taht a randomly-selected day of the year (from the 366 possible days) is in April?


2- What is the probability that a card selected from a deck is an ace or a heart?

3- What is the probability that a five-card poker hand contains a flush, that is, five cards at the same suit?

4- Which is more likely, rolling a total of 8 when two dice are rolled, or rolling a total of 8 when three dice are rolled?


please help me as fast as possible

thanks & regards

[[Digital-X-Treme]]

Question 1. As there can be 'normal' years and 'leap' years, i will consider both.

In a 'normal' year, there are 365 days, spread throughout the months as follows:

Jan 31
Feb 28
Mar 31
Apr 30
May 31
Jun 30
Jul 31
Aug 31
Sep 30
Oct 31
Nov 30
Dec 31

The probability of a randomly selected day of the year being in April is 30/365 = 6/73.

In a 'leap' year, Feb has an extra day. Therefore the probability becomes 30/366 = 5/61.

Question 2. Assuming a full deck of 52 cards. P(A) = Ace, P(H) = Heart.

P(A) = 4/52
P(H) = 13/52
P(A n H) = 1/52 (Probability of A intersection H, both an Ace and a Heart)
P(A U H) = P(A) + P(H) - P(A n H) = (4/52) + (13/52) - (1/52) = (4/13)

So the probability that a card selected from a deck is an ace or a heart is 4/13.

Questoin 3.
Do you need to take other players' hands into consideration, or do you just want to know what the probability will be if the 5 cards are drawn straight from a deck of 52?

Question 4.
If you have two dice, the possible totals of their scores when thrown together can be summarised as follows (this table probably wont show up right... )

2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12


There are 36 possible combinations of scores, 5 of which total 8. Therefore the probability of scoring an eight with two dice is 5/36 (0.1389 4 d.p.).

Three dice gets a bit trickier. You will basically be dealing with a 3-dimensional array of values (as opossed to the two dimensional array above). Here are the possible totals of their scores:

3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
8 9 10 11 12 13


4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
8 9 10 11 12 13
9 10 11 12 13 14


5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
8 9 10 11 12 13
9 10 11 12 13 14
10 11 12 13 14 15


6 7 8 9 10 11
7 8 9 10 11 12
8 9 10 11 12 13
9 10 11 12 13 14
10 11 12 13 14 15
11 12 13 14 15 16


7 8 9 10 11 12
8 9 10 11 12 13
9 10 11 12 13 14
10 11 12 13 14 15
11 12 13 14 15 16
12 13 14 15 16 17


8 9 10 11 12 13
9 10 11 12 13 14
10 11 12 13 14 15
11 12 13 14 15 16
12 13 14 15 16 17
13 14 15 16 17 18

There are 216 possible combinations of scores, 21 of which total 8. Therefore the probability of scoring an eight with three dice is 21/216 = 7/72 (0.0972 4 d.p.).

Therefore, you are more likely to roll a score of 8 when two dice are thrown.

Hope this helps. Fell free to ***** slap if any of this is wrong etc...

Later

[proff.hacker]

Thank you [Digital-X-Treme]

I have to say 'Thank you' million of times to you.


for Question 3:
I dont know because the teacher gave me the question as is and he didn't give any more clarification.

so if u can answer according to both ways or the way u think it is.

[riis]

3- What is the probability that a five-card poker hand contains a flush, that is, five cards at the same suit?

I don't think that there's any difference. To you, it is unknown which cards are in the hands of the other players or in the deck.

The probability is: 12/51 * 11/50 * 10/49 * 9/48 = 11880 / 5997600 = 0.001981

(Edited this post since I saw + signs, instead of * signs!)

[[Digital-X-Treme]]

Ok, assuming you want the probability of having 5 cards of the same suite, you would calculate it as follows:

(13/52) * (12/51) * (11/50) * (10/49) * (9/48) = 33 / 66640.

Make sure you understand the stuff above... its easy to post and get help, but there will come a time when you might not be able to get help...

[proff.hacker]

There are 36 possible combinations of scores, 5 of which total 8. Therefore the probability of scoring an eight with two dice is 5/36 (0.1389 4 d.p.).

There are 216 possible combinations of scores, 21 of which total 8. Therefore the probability of scoring an eight with three dice is 21/216 = 7/72 (0.0972 4 d.p.).

thanks Mr [Digital-X-Treme]

i understood your answers except that 5 and 21 you got.

how did you get it?

thanks

[[Digital-X-Treme]]

You are dealing with a question on 'Probability Distributions'. A probability distribution is the set of all values of a random variable together with their associated probabilities.

Consider the first dice problem, where you have two dice. Let 's' represent the sum of their scores when thrown. The possible combinations of scores are as follows:

2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12

From this, we can see that:

There is a 1/36 chance of scoring two.
There is a 2/36 chance of scoring three.
There is a 3/36 chance of scoring four.
There is a 4/36 chance of scoring five.
There is a 5/36 chance of scoring six.
There is a 6/36 chance of scoring seven.
There is a 5/36 chance of scoring eight.
There is a 4/36 chance of scoring nine.
There is a 3/36 chance of scoring ten.
There is a 2/36 chance of scoring eleven.
There is a 1/36 chance of scoring twelve.


We can now construct a probability distribution of this information... All probabilities add up to 1.

s P(S = s)
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36

You can now find information such as P(s), which is "what is the probability of scoring 's'?". For example, as your question asked, p(8) (probability of scoring an eight) = 5/36.

The 3 dice problem is an exension to this, with a larger number of possible scores.

Hope this helps.

[riis]

Originally posted by [Digital-X-Treme]
Ok, assuming you want the probability of having 5 cards of the same suite, you would calculate it as follows:

(13/52) * (12/51) * (11/50) * (10/49) * (9/48) = 33 / 66640.

This is only true if you need 5 cards of a particular suite (e.g. hearts). If the suite is irrelevant (which I assumed implicitely in my previous post), the first factor is superfluous.

[[Digital-X-Treme]]

Originally posted by riis

This is only true if you need 5 cards of a particular suit (e.g. hearts).

He specified 5 cards of the same suit.

Originally posted by riis

If the suite is irrelevant (which I assumed implicitely in my previous post), the first factor is superfluous.

The suit isn't irrelevant, as proff.hacker specified, 5 cards of the same suit.

(13/52) * (12/51) * (11/50) * (10/49) * (9/48) = 33 / 66640.

This is the same for all 4 suits. Think about it like this.

I have 52 cards. I have a 13/52 chance of getting a card from a particular suit.

I now have 51 cards. I have a 12/51 chance of getting a card from the same suit as above...

I now have 50 cards...

[riis]

Originally posted by [Digital-X-Treme]


(13/52) * (12/51) * (11/50) * (10/49) * (9/48) = 33 / 66640.

This is the same for all 4 suits. Think about it like this.

I know what you're trying to explain, Digital-X-Treme. What I was saying is:

4 * ((13/52) * (12/51) * (11/50) * (10/49) * (9/48)) = (52/52) * (12/51) * (11/50) * (10/49) * (9/48) = (12/51) * (11/50) * (10/49) * (9/48)

The first card you draw determines of which suit your flush will be.