probability :chips and their values

amal_

New member
Joined
Sep 20, 2021
Messages
9
Hello ,I hope you are doing great !I am having a problem with the operations I should use for the following exercise .
" A box contains 10 chips, of which four are red, three are white, two are blue and one is yellow. Red, white, blue and yellow chips are worth $1, $2, $5 and $10, respectively, Two chips are selected at random without replacement from the box. Find the probability that the selected chips are(a) worth at least $5;(b) worth at most $10."

I think it's easier to calculate the probability of the opposite events .For the first question ,selecting chips that are worth less than 5 dollars is equivalent to selecting (2 red chips),(1 red chip and 1 white one)or(2 white ones )however I am confused about the operations that i should use to calculate these events .Selecting 1 red chip and 1 white one is it equivalent to (4/10)*(3/9) or is it equivalent to (4C1 *3C1) /(10C2).
I would very much appreciate it if you could suggest me any courses or videos about probability that could help me with understand the course better .Thank you very much !
 
Selecting 1 red chip and 1 white one is it equivalent to (4/10)*(3/9) or is it equivalent to (4C1 *3C1) /(10C2).
The first is the probability of selecting RW in that order; you can double it to get the correct answer (allowing for both orders). The second will give the same result without modification.

I think it's easier to calculate the probability of the opposite events .For the first question ,selecting chips that are worth less than 5 dollars is equivalent to selecting (2 red chips),(1 red chip and 1 white one)or(2 white ones )
I find it helpful to put all possible pairs into a 4x4 table like this:

R (1)W (2)B (5)Y (10)
R (1)23611
W (2)34712
B (5)671015
Y (10)11121520

I've marked the two events you want to calculate (the complements) in red and green. Looking at them, I see a somewhat easier way to calculate each of them, all at once, than the sums you require if you continue as you are going. This assumes you have learned about "and" and "or".

But what you're doing can be fine.
 
The first is the probability of selecting RW in that order; you can double it to get the correct answer (allowing for both orders). The second will give the same result without modification.


I find it helpful to put all possible pairs into a 4x4 table like this:

R (1)W (2)B (5)Y (10)
R (1)23611
W (2)34712
B (5)671015
Y (10)11121520

I've marked the two events you want to calculate (the complements) in red and green. Looking at them, I see a somewhat easier way to calculate each of them, all at once, than the sums you require if you continue as you are going. This assumes you have learned about "and" and "or".

But what you're doing can be fine.
Thank you so much for your time! This has been extremely useful.
 
I would list all possible RVs for 2 colors first and treating this as a boy/girl debate team problem. rr, ww, bb plus a c(4,2) = 9 possible RVs. For each RV just apply the debate team formula, for instance a wy event has a prob: (c(4,0)*c(3,1)*c(2,0)*c(1,1))/c(10,2) and ww event has a prob:
(c(4,0)*c(3,2)*c(2,0)*c(1,0))/c(10,2)

All 9 possible RVs for 2 colors have different sum amounts as: $2, $4, $10, $3, $6, $11, $7, $12, $15 so each Sum Amount is unique and reflects the associated RV for 2 colors. So last step just sum the probabilities of each RV satisfying the question requirement.

what do you guys think?
 
Top