Probablity Help

[ElephantStew]

Gah....i hate probability  . and some help would be much appreciated

1) a random variable is such that its mean is 4 and its standard deviation is 3. Find what is.

i got this far: , , therefore .




i'm not sure if this right...

2) Suppose there are two boxes, box A and box B. Box A contains 4 red balls and 6 black balls. Box B contains 3 red and 7 black balls. Suppose tht Lauren selects a box at random, and then a ball t random from the box.
Find the probability that Lauren chooses a red ball.
Find the probability that Lauren chose box A given that she chooses a red ball.

i know the second one is quite simple, but thats besides the point coz i have nfi  :uglystupid2:

[Collin Li]

1) a random variable is such that its mean is 4 and its standard deviation is 3. Find what is.

i got this far: , , therefore .




i'm not sure if this right...

Right idea, but you didn't substitute it right. I'm glad you didn't try to do anything silly like instead, because that does not equal in general.

[ElephantStew]

ahhh k,  ty

[Collin Li]

Let R be the event that Lauren chooses a red ball:



Let A be the event that Lauren chooses Box A:

[ElephantStew]

okay, i need help with a question:

x           0     1     2
Pr(X=x)   0.5   a     b

Find the values of a and b, if the mean of X is 0.7

i got this far:
a + 2b = 0.7
then id use simultaneous equations to find the values of a and b
the real question is....am i on the right track

[Collin Li]

Yes, you are on the right track. Here is the solution:

For a discrete random probability distribution:


(1)

Mean:


(2)

Subtracting the first equation from the second equation yields:

[excal]

Yes, you are on the right track. Here is the solution:

For a discrete random probability distribution:


(1)

Mean:


(2)

Subtracting the first equation from the second equation yields:

I can confirm this solution is correct.

[Glockmeister]

Yes, you are on the right track. Here is the solution:

For a discrete random probability distribution:


(1)

Mean:


(2)

Subtracting the first equation from the second equation yields:

I can confirm this solution is correct.

me three

[ElephantStew]

gah need more help

okay...
1) Steve devises the following game. He will toss two coins, and if both are heads he will pay his friend $1.00, otherwise his friend must pay him 50c. How much will Steve expect to win or lose per came in the long run using this strategy?

i got this far....

Pr(HH)  = 1/4
Pr(Not HH) = 3/4

then i used a probability table thing

x             0.5     1
Pr(X=x)    3/4     1/4
Then i figured out E(X) to be 5/8
now im stumped, where do i go from here (uis this even how you do it?)

2)At a particular university the probability that a randomly selected student studying a business degree is 0.5, that they are male is 0.5, and that they are male business student is 0.3. what is the probability that a randomly selecte student is either male or a business student or both?

Would i be correct in assuming that i use a Karnaugh table?

btw...thanks for your help in the previous questions.

[ed_saifa]

The tells that he is expected to win cents

2) i would use those tables as well

[Mao]

gah need more help

okay...
1) Steve devises the following game. He will toss two coins, and if both are heads he will pay his friend $1.00, otherwise his friend must pay him 50c. How much will Steve expect to win or lose per came in the long run using this strategy?

i got this far....

Pr(HH)  = 1/4
Pr(Not HH) = 3/4

then i used a probability table thing

then:

HH = -$1.00

HH' = $0.50

win

note that the $1.00 is a loss to him, hence be negative. your did not take that into consideration.

2)At a particular university the probability that a randomly selected student studying a business degree is 0.5, that they are male is 0.5, and that they are male business student is 0.3. what is the probability that a randomly selecte student is either male or a business student or both?

Would i be correct in assuming that i use a Karnaugh table?

btw...thanks for your help in the previous questions.

there is no need. the formula you need is

[ElephantStew]

note that the $1.00 is a loss to him, hence be negative. your did not take that into consideration.

gah fuck probability
thanks a lot mao

[ElephantStew]

could someone please tell me what markov and bernoulli sequences are???
thanks

[cara.mel]

Markov sequence is like your Q2 in the other thread I just replied to

Bernoulli is like 'I chuck a coin 20 times, whats the probability I get 13 heads'

[dcc]

Bernoulli is like 'I chuck a coin 20 times, whats the probability I get 13 heads'

Let X be the number of heads in total from chucking a coin 20 times.