Question

Single cards, chosen at random, are given away with bars of chocolate. Each card shows a picture of one of 20 different football players. Richard needs just one picture to complete his collection. He buys 5 bars of chocolate and looks at all the pictures. Find the probability that
(i) Richard does not complete his collection
(ii) he has the required picture exactly once,
(iii) he completes his collection with the third picture he looks at.

Workings and explaination too!

Answer 1

the probability that he gets i on any one card is one out of twenty or \(\frac{1}{20}\) or \(.05\)
clear so far?

Answer 2

*gets IT

Answer 3

the probability that he does not get in on any one card is therefore
\[\frac{19}{20}\] or \(0.95\)
the probability that he does not get it five times in a row is \(0.95^5\)

Answer 4

Okay what about they rest?

Answer 5

you familiar with the "binomial" distribution? we can reason it out anyway, but that is what we are using

Answer 6

Yes

Answer 7

ok then with no explanation at all (no reasoning) we set it up as
\[p=.05,(1-p)=.95, n=5, P(x=1)=\binom{5}{1}.05\times .95^4\]

Answer 8

:( looks like I need reasoning. Sorry..

Answer 9

k no problem, it makes sense with reasoning and is just as easy
he picks 5 cards,
one has the one he is looking for
four to not

Answer 10

the probability of getting say (S, F, F, F, F) where as usual S is success and F is failure (getting the card on the first pick not on the last four) is therefore
\[.05\times .95\times .95\times .95\times .95\] more succinctly written as
\[.05\times .95^4\]
we multiply because the events are independent

Answer 11

but there are five different ways for him to get the card he he is looking for
(S, F, F, F, F)
(F, S, F, F, F)
(F, F, S, F, F)
(F, F, F, S, F)
(F, F, F, F, S)

Answer 12

since these events all have the same probability of \(.05\times .95^4\) and since they are mutually exclusive (can't both happen at the same time) we add up the probabilities, i.e. multiply
\(.05\times .95^4\)times \(5\) and get
\[P(x=1)=5\times .05\times .95^4\]

Answer 13

that was a lot of words, but i didn't want to skip any detail
hope it is more or less clear

Answer 14

I see that. Okay man thanks so much, you're really good in explaining. Next questions!

Answer 15

question*

Answer 16

btw before we get to the next question, that should explain why if you have a bernoulli trials (two possible outcomes, S, F independent ) with \(p\) as the probability of success and \(1-p\) of failure, then the probability you get exactly \(k\) successes in \(n \) trials is
\[P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}\]

Answer 17

third one is easier
doesn't get it
doesn't get it
does goes it

\(.95\times .95\times .05\)

Answer 18

Oh cool! Got it. Erm, referring to ii), the answer scheme says 5c1 why is that so?

Answer 19

that is the thing i wrote as \(\binom{5}{1}\) which is just the 5 from the 5 ways to get it

Answer 20

\(\binom{5}{1}\) is the adult version of \(_5C_1\) but it is just the number 5 in any case

Answer 21

Yeah, I know. BUT why is it 5 choose 1

Answer 22

remember my long answer above? i actually listed the 5 ways to get one success and four failures

Answer 23

Ohhh. Okayy thanks a bunch! You re really good! :)

Answer 24

yw