2 really hard, long probability problems...please help!?!?!?

Question

when dealing with a problem with probabilities, how do you know to whether multiply or add the probabilities??? i heard that you multiply when "and" is in the answer or add when "or" is in the answer...or it might be vice versa (please explain this to me also!).

here is an example if it didn't make sense (please explain):

1.) A.) there are 5 blue, 4 red, 3 yellow, and 2 black marbles in a bag. If two marbles are picked at random (without replacement) what is the probability that both selections (first AND second) are red?
B.) same problem as 1. A but this time two marbles are selected at random WITH REPLACEMENT

2.) A.) there are 5 blue, 4 red, 3 yellow, and 2 black marbles. If two marbles are picked at random (without replacement) what is the probability that either selections (OR) is red?
B.) same problem as 2 A, but this time WITH replacement.

short:
-do all problems
-explain whether multiply when it says "and" or add when it says "or"..or vice versa
-explain why so
10 points!!!

Answer 1

(1) You have 14 marbles, 4 of which are red. Therefore your chance of getting a red marble on the first pick is 4 / 14. Assuming success, you now have 13 marbles, 3 of which are red. The chance of getting another red marble is 3 / 13. So the probability of drawing two red marbles in a row is

( 4 / 14 ) ( 3 / 13 ) = 12 / 182 = 6 / 91

(B) Using similar reasoning,

( 4 / 14 ) ( 4 / 14 ) = 4 / 41

(2) The probability of drawing two NON red marbles in a row, without replacement, is

( 10 / 14 ) ( 10 / 13 ) = 50 / 91

so the probability of getting at least one red is 41 / 91

(B) The probability of getting two non reds is

( 10 / 14 ) ( 10 / 14 ) = 25 / 49

so the probability of getting at least one red is 24 / 49.

Answer 2

There is no hard and fast rule to add or multiply probabilities.

If events A and B are INDEPENDENT events then:

P(A ∩ B) = P(A) * P(B)

If A and B are any two events, independent or not, the following is always true:

P(A U B) = P(A) + P(B) - P(A ∩ B)

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1a. This is solved in two ways:

Forgive the abuse of notation:

P(1st draw in red) = 4/14
P(2nd draw is red | 1st draw is red) = 3/13

P(Both draws are red) = 4/14 * 3/13 = 0.06593407

the other way to solve this is Let X be the number of Red marbles selected from the bag. X has the Hyper Geometric Distribution.

K = number of items to be drawn
N = total objects
M = number of objects of a given type

The probability mass function for the hypergeometric distribution is defined as:

P(X = x | N, M, K) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K ) for x = {0, 1, 2, 3, ..., K}
P(X = 0 | N, M, K) = 0 otherwise

If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr

the N C K is the total number of possible combinations of K objects drawn from N objects.
the M C x is the number of combinations of getting x objects of the given type
the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn.

P(X = 0 ) = 0.4945055
P(X = 1 ) = 0.4395604
P(X = 2 ) = 0.06593407

--- ---- ---- ---- ---- ---- ----

1b. again, two ways to do this

when you are drawing with replacement now the probability of getting a red marble on either draw is 4/14

P(Both Red) = 4/14 * 4/14 = 0.08163265

Let X be the number of red marbles drawn. X has the binomial distribution with n = 2 trials and success probability p = 0.2857143 = 4/14.

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n , p )

the mean of the binomial distribution is n * p = 0.5714286
the variance of the binomial distribution is n * p * (1 - p) = 0.4081633
the standard deviation is the square root of the variance = 0.6388766

The Probability Mass Function, PDF,
f(X) = P(X = x) is:

P(X = 0 ) = 0.510204081632653
P(X = 1 ) = 0.4081632653061226
P(X = 2 ) = 0.0816326530612245

--- --- --- --- --- ---

2a.

Let R be the event that draw is Red
Let N be the event the draw is not Red

P(either draw is red) = 1 - P(no draw is red)

P(no draw is red) = P(NN) = 10/14 * 9/13
= 0.4945055

P(either draw is red) = 1 - 0.4945055 = 0.5054945


Let X be the number of red marbles. X has the hypergeometric distribution. Find P(X ≥ 1)

P(X ≥ 1) = P(X = 1) + P(X = 2)
= 0.4395604 + 0.06593407
= 0.5054945
--- --- --- -- --- ---

2b.

P(A) = 4/14
P(B) = 4/14
P(A ∩ B) = 16/196

P(A U B) = P(A) + P(B) - P(A ∩ B)
P(A U B) = 4/14 + 4/14 - 16/196
P(A U B) = 0.4897959

Let X be the number of red marbles drawn. X has the binomial distribution. Find P(X ≥ 1)

P(X ≥ 1) = P(X = 1) + P(X = 2)
P(X ≥ 1) = 0.4081632653061226 + 0.0816326530612245
= 0.4897959

Answer 3

You add the probabilities when you are dealing with mutually exclusive events. You multiply the probabilities when the events are conditional upon one another.

For example, if I asked you what would be the probability of getting a head OR a tail if I tossed one coin one time, you would add the probability of a head 0.5 to that of a head 0.5 to get 1.0. These events are mutually exclusive. You can't get a head AND a tail at the same time, so the probability of getting one to that of the probability of getting the other .

If I asked you what was the probability of getting two heads in succession if you tossed the same coin twice, to succeed, the probability of getting the second head is CONDITIONAL on having gotten the first head. Therefore, you multiply the 2 probabilities together. 0.5 X 0.5 = 0.25

I hope this helps you to get a better understanding.

1) A - Here is an example of CONDITIONAL probabilities. There are 14 marbles in all, 4 of which are red.

The probability of selecting 2 reds in succession (just like the 2 heads example above, but here, we don't replace the first marble) is

4/14 X 3/13 = 12/182

B) with replacement:

4/14 X 4/14 = 16/196

I'm going to let you do the last question on your own. Just follow the logic above, and you'll do fine.

Answer 4

I suspect my argument is not going to be well received. I say the probability is 50% Let a ∈ℝ, Let b ∈ℝ and randomly select the values for a and b. As already noted, for a ≤ 0, P( a < b²) = 1, this is trivial. Only slightly less trivial is the idea that P(a < 0 ) = 1/2 and thus P( a < b² | a ≤ 0) = 1 and P( a < b² ) ≥ 1/2 Now consider what happens when a > 0 For a > 0, while it is easy to show there is a non zero probability for a finite b, the limit, the probability is zero. a < b² is equivalent to saying 0 < a < b², remember we are only looking at a > 0. If this a finite interval on an infinite line. The probability that a is an element of this interval is zero. P( a < b² | a > 0) = 0 As such we have a total probability P( a < b² ) = P( a < b² | a ≤ 0) * P(a ≤ 0) + P( a < b² | a > 0) * P(a > 0) = 1 * 1/2 + 0 * 1/2 = 1/2 Remember, this is because of the infinite sets. No matter what type of interval you draw on paper or on a computer you will find a finite probability that appears to approach 1. But this is due to the finite random number generators on the computer and if we had this question asked with finite values there would be a a solution greater than 50%. I don't mean to be condescending, but please explain why using the Gaussian to approximate a uniform distribution is a good idea? Aren't infinite numbers fun. Cantor when mad working with them! :)

Answer 5

1a) you start with a total of 14 marbles, of which 4 are red, the probability of choosing a red marble on the first attempt is 4/14; if you choose a red marble, then you have 3 red marbles out of 13 remaining, so the probability of choosing a red marble on the second attempt is 3/13

The JOINT probability of two events occuring is the prob of the first times the prob of the second, so the prob of choosing two red marbles in a row is 4/14 * 3/13

b) with replacement, the problem is conceptually easier; the prob of choosing a red marble on the first attempt is 4/14; now, if you replace the red marble you are back to 4 red marbles out of a total of 14, and the probability of the second choice is also 4/14, so the probability of the two events is now the product of 4/14*4/14

Answer 6

I don't know how to solve these problems, but I can explain what WITH REPLACEMENT and WITHOUT REPLACEMENT means.

WITH replacement- for example, lets say you have 3 marbles. you take one out of the jar. you put it back in. you still have three marbles.

3-1+1=3

WITHOUT replacement- for example, lets say you have 5 marbles. you take one out. you DONT put in back in the jar. you now have four marbles.

5-1=4