Probability Questions Part 2 HELLLLPP!?

Question

A pair of dice are rolled. FInd the following:

1. Odds in FAVOR of doubles

2. The probability of a sum greater than 3

3. The probability of a sum that is odd and less than 4.



Two cards are drawn , without replacement from a standard 52-card deck. Find the probability of the following events:

1. both cards are black

2. the first card is a face card and the second is red.

Answer 1

1. There are only 6 doubles: (1,1) , (2,2) , (3,3) ,... ,(6,6) each with probability of occurrence: 1/36.
Thus the total probability is 6/36 = 1/6. The odds will be 1:5.

2. There are two sums not greater than 3: namely 2 & 3.
1 - P(x=2) - P(x=3) = 1 - 1/36 - 2/36 = 33/36.

3. P(x=3) = 2/36 ... the only odd sum less than 4 is 3.


©
1. (26*25)/(52*51)

2. 1st card:Red face & 2nd: red: (6/52)*(25/51)
1st card:Black face & 2nd: red: (6/52)*(26/51)
Probability: (6/52)*(25/51) + (6/52)*(26/51)

Answer 2

I presume you want the answers, which have been given by others. But would you also like to know how to work this kind of problems on your own?

For the dice problems, write out a 6 X 6 table, each row 1...6 represents a dot on the red dice, each column 1...6 is a dot on the green one. (I give them color for easy ID).

So how many possible outcomes (N) are there when tossing two dice? That's 6 X 6, because there are that many elements in a 6 X 6 table. Check this out:

...1..2..3..4...5..6
1..d.x..x...x...x..x
2..x..d..x..x...x..x
3..x...x..d..x..x..x
4..and so on.....
5
6

Probability is defined by P(s) = n(s)/N; where n(s) is the number of possible successes and N is the total number of possible outcomes for one experiment (e.g., one toss of the two dice). So what's the probability of a double P(s = double) = n(s = double)/N = 6/36 = 1/6. This follows from the table because there are n(s = double) = 6 ways to get a double and that's on the diagonal of the table, which I marked with the "d" in each row/column.

How about P(s > 3) = n(s > 3)/N. How many ways are there, from the table, to get 4 or more as a sum? In the first row there are four ways: 1:3, 1:4, 1:5, and 1:6. In the second row, we have five ways: 2:2, 2:3, etc. Just add up all the ways and you have n(s > 3) = n(s >= 4). N still is 36 because that represents all possible outcome, including failures, which are s <= 3.

And that brings us to another point, finding n(s <= 3) is quicker than finding n(s > 3) because there are fewer combinations to count. By looking at the table, we can quickly see there are but n(s <= 3) = 3 ways to roll a three or two. So taking advantage of the fact that the probability of your dice coming up with some number between 2 and 12 is P(any) = n(any)/N = 36/36 = 1.00, we can write P(any) = P(s > 3) + P(s <= 3) = 1.0 and then, ta da, P(s > 3) = 1.0 - P(s <= 3) = 1.0 - 3/36 = 11/12.

For 3. in your dice question. Count up the number of sums less than 4 and odd. n(s < 4 and odd) is that number. So P(s < and odd) = n(s < 4 and odd)/N, just like we did the other two questions.

The card questions are less visual because 52 cards are a lot to visualize. But working these questions is still a matter of seeing what you have.

For example, what is the total number of possible outcomes (success and failure) when drawing a single card from a full deck. That's N = 52, you could draw any one card from the entire deck. How about on the next draw if you don't put the first card back into the deck? That's N - 1 = 51 because one card from the first draw is missing. So now we have the total number of possible outcomes for the first and then the second draw without replacement. We're half way there.

Now lets find the n(s) numbers. How many black cards are in a full deck? n(black|1) = 26 for spades and clubs on the first draw. So with the virgin deck, we have P(black|1) = n(black|1)/N = 26/52 = 1/2.

The fun part begins with the second draw. What's the chances of drawing two black cards if the first one is not a black card...nada, zilch, zero. You can't draw two consecutive blacks if the first one isn't black. So, for that second draw n(black|2) = n(black|1) - 1 = 25. Which means drawing a black on the second pull is P(black|2) = n(black|2)/N = 25/51.

Now we combine the two probabilities to get the joint probability of drawing two consecutive blacks. That's simply the product of the two probabilities for the two independent events (the two draws). Thus we have P(black|1 and black|2) = P(black|1)P(black|2) = 26*25/52*51 for that.

You can do 2. similarly. P(face) = n(face)/N and P(red) = n(red)/N-1. n(face) is easy, it's a virgin deck with 12 face cards. The second probability is more problematic. Clearly that face card we drew could be either red or black. If it's red then n(red) = 25 for the second draw. If it was black, then n(red) = 26.

This brings us to something called conditional probability. The condition being what color was the face card drawn on the first draw? The probability of drawing a red is the sum of the two conditions because either outcome is possible. Thus, P(red) = P(red|red) + P(red|black) where P(red|*) is the probability of drawing a red card given the condition * from the first draw.

But still the joint probability is just the multiple of the two draw probabilities. So we have P(face and red) = P(face)P(red) = P(face)[P(red|red) + P(red|black)] and you can do the math.

Answer 3

1. 9/36 so 1:3
2. 33/36
3. 2/36



1. (2C26)/(2C52)
2. (12)/(52) * (26/51)

though that 26 could be a 25 if your face card was red but close enough


I THINK SOME PEOPLE FORGOT THAT THERE ARE 26 RED CARDS (DIAMONDS AND HEARTS) AND 26 BLACK CARDS (CLUBS AND SPADES) EVEN THOUGH THEIR ANSWER(S) IS OTHERWISE CORRECT

Answer 4

1. P(odds) = 9/36
P(odds) = 1/3


2. P(s>3) = 33/36 ..[neglecting three cases - (1,1),(1,2),(2,1)]
P(s>3) = 11/12

3. P(o<4) = 2/36 [(1,2),(2,1)]
P(o<4) = 1/18





1. P(both black) = 13/52 * 12/51
P(both black) = 4*4/17
P(both black) = 16/17

2. P(ace and red) = 4/52*13/51
P(ace and red) = 1/51