Question

A hat contains 25 coins - 7 gold, 6 silver and 12 copper. We randomly select two coins from the hat without replacement.

(a) What is the probability that both selected coins are the same colour?

(b) Let X be the number of gold coins selected. Find the probability distribution of X.

(c) What is the expected value of X?

(d) What is the variance of X?

(e) What is the probability that X = 1 if no silver coins are selected?

Answer 1

Hey, so what have you done so far?

Answer 2

for a) P(( G1 ∩ G2 ) ∪ (S1 ∩ S2) ∪ (C1 ∩ C2))
but idk how to go after this

Answer 3

b) x 0 1 2
P(X=x) 0.2435 0.4695 0.2504

Answer 4

Okay, break it down a little. \(P(G1)=7/25\) and \(P(G2)=6/24\) so \(P(G1∩G2)=\frac{7*6}{25*24}=\frac{42}{6000}=0.07\). Can you work this out for S and C?

Answer 5

okay let me do this! can u please be here while i try to solve it?

Answer 6

I'll be around for another 10 minutes or so, we should be able to get through it in that.

Answer 7

P(S1)= 6/25 and P(S2)= 5/24 ; P(S1∩S2)= (6x5)/(25x24)=0.005

P(C1)= 12/25 and P(C2)= 11/24; P(C1∩C2)= (12x11)/(25x24)=0.022

Answer 8

Yep, well done! So all we do now is sum those for our answer to (a).

Answer 9

okay! thanks! so it shud be: 0.097 (sum of all three)

Answer 10

Perfect. For (b), how did you get those numbers? They're wrong as they should add up to 1.

Answer 11

i cant remember how i got it. can u give me a clue on how to do (b)

Answer 12

So we have:
\(P(X=0)=\frac{18}{25}*\frac{17}{24}=0.51\)
\(P(X=1)=\frac{7}{25}*\frac{18}{24}+\frac{18}{25}*\frac{7}{24}=0.42\)
\(P(X=2)=\frac{7}{25}*\frac{6}{24}=0.07\)

Answer 13

Does that make sense? Remember X is the number of Gold coins selected in two draws.

Answer 14

why did u use 18,17 for x=0? initially isnt there only 7 gold coins, so 7 x 2 = 14

Answer 15

\(P(X=0)\) means the probability of getting 0 gold coins, so for the first draw we have a 18/25 chance and for the second, we have a 17/24 chance. Does that make sense?

Answer 16

okay! yes, cos u take 25-7 (since theres not gold coins selected), and for PX=1, why must u multiply 7 with 18

Answer 17

To get only 1 gold coin, we either have (gold + not gold) or (not gold + gold). Does that make sense now?

Answer 18

yes very good explanation! perfect. c) shud X=1? since it has to add up to 1

Answer 19

No, the expected value is the sum of \(x_i*P(x_i)\). So \(0*0.51+1*0.42+2*0.07=0.56\)

Answer 20

is expected value also the mean?

Answer 21

Yes :)

Answer 22

d) var(X) = total of P(x-E(X))^2, where do i get the values to plugin

Answer 23

\[var(x)=\sum(x^2*P(x))-\mu^2\]\[var(x)=0^2*\frac{51}{100}+1*\frac{21}{50}+2^2*\frac{7}{100}-0.56=\frac{483}{1250}\]

Answer 24

Then for (e), it's just the probability of (gold and copper) or (copper and gold). GOtta go now, good luck!