My Math Forum Problems with P.M.F. and probability.

 March 2nd, 2018, 09:49 PM #1 Newbie   Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2 Problems with P.M.F. and probability. I am having some troubles with this question. Question 1: I) Suppose Lucky 6 scratch tickets all have an independent probability of 0.05 of being a winner. Kathy will keep buying scratch tickets until she (a) gets a winner or (b) she has scratched 20 tickets, which ever comes first. Let X be the number of scratch tickets she buys. Find the p.m.f. of X. II) Using the same set up as question 1, what is the probability she buys less than 17 scratch tickets? Simplify your answer. Any help would be appreciated. Thanks.
 March 2nd, 2018, 10:26 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,976 Thanks: 1026 $p = P[\text{win}] = 1-P[\text{loss}] = 1 - (1-0.05)^6 = 0.264908109375$ $P[\text{win at kth ticket}] = (1-p)^{k-1}p$ so $P_X(k) = \begin{cases}0 &k<1 \\(1-p)^{k-1}p &1\leq k < 20\\ 1-\sum \limits_{k=1}^{19} (1-p)^{k-1}p &k=20\\0 &20 < k \end{cases}$ $20$ is oddball because she picks 20 if she wins on 20 and if she never wins. given this you can figure out (II) Thanks from Shanonhaliwell
March 3rd, 2018, 09:43 AM   #3
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Quote:
 Originally Posted by romsek $p = P[\text{win}] = 1-P[\text{loss}] = 1 - (1-0.05)^6 = 0.264908109375$ $P[\text{win at kth ticket}] = (1-p)^{k-1}p$ so $P_X(k) = \begin{cases}0 &k<1 \\(1-p)^{k-1}p &1\leq k < 20\\ 1-\sum \limits_{k=1}^{19} (1-p)^{k-1}p &k=20\\0 &20 < k \end{cases}$ $20$ is oddball because she picks 20 if she wins on 20 and if she never wins. given this you can figure out (II)
Thank you so much.

I assume X be the number of tickets until winning.
X=k
Part II)
P_X(x<17) = P_X(1)+P_X(2)+P_X(3)+.......+P_X(15)+P_X(16)=0.729 7486

However, I am still being confused your result in part I.
I don't know if I understand the P[loss] correctly, but I think the P[loss] is the probability of all six parts in a Lucky 6 scratch ticket being loss which is (1-0.05)^6. Am I right?
The most parts that I don't understand are the p.m.f of the case (1<=k<20) and the p.m.f. of the case (k=20). Why do we have to separate these 2 cases. Can we have the same p.m.f. when k=20 which is (1-p)^19*p.
I put this into R-console then it seems like the probability of (1-p)^19*p would be different than if I use your p.m.f. I don't know why.

March 3rd, 2018, 11:21 AM   #4
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Quote:
 Originally Posted by Shanonhaliwell However, I am still being confused your result in part I. I don't know if I understand the P[loss] correctly, but I think the P[loss] is the probability of all six parts in a Lucky 6 scratch ticket being loss which is (1-0.05)^6. Am I right?
Yes. You lose if none of the 6 tickets are chosen. They are independent so this is $(1-0.05)^6$

Then we subtract that from 1 to get the probability that a given bought ticket was one of the winning 6 tickets.

Quote:
 The most parts that I don't understand are the p.m.f of the case (1<=k<20) and the p.m.f. of the case (k=20). Why do we have to separate these 2 cases. Can we have the same p.m.f. when k=20 which is (1-p)^19*p. I put this into R-console then it seems like the probability of (1-p)^19*p would be different than if I use your p.m.f. I don't know why.
Two things can happen when she buys the 20th ticket. She can win and stop. The probability of this is $p(1-p)^{19}$

or she can lose.. and stop because she's not buying any more than 20 tickets.

so for example

$P[19] = P[\text{bought winning ticket on 19th try}]$

but

$P[20] = P[\text{bought winning ticket on 20th try}] + P[\text{did not buy any winning tickets in 20 tries}]$

$P[\text{did not buy any winning tickets in 20 tries}] = (1-p)^{20}$

so

$P[20]=1-\sum \limits_{k=1}^{19}p(1-p)^{k-1} = p(1-p)^{19} + (1-p)^{20}$

come to think of it the second form is much cleaner so I would write

$P_X(k) = \begin{cases}0 &k<1 \\p(1-p)^{k-1} &1\leq k \leq 19\\p(1-p)^{19}+(1-p)^{20} &k=20\\0 &20 < k \end{cases}$

Last edited by romsek; March 3rd, 2018 at 11:25 AM.

March 3rd, 2018, 04:37 PM   #5
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Quote:
 Originally Posted by romsek Yes. You lose if none of the 6 tickets are chosen. They are independent so this is $(1-0.05)^6$ Then we subtract that from 1 to get the probability that a given bought ticket was one of the winning 6 tickets. Two things can happen when she buys the 20th ticket. She can win and stop. The probability of this is $p(1-p)^{19}$ or she can lose.. and stop because she's not buying any more than 20 tickets. so for example $P[19] = P[\text{bought winning ticket on 19th try}]$ but $P[20] = P[\text{bought winning ticket on 20th try}] + P[\text{did not buy any winning tickets in 20 tries}]$ $P[\text{did not buy any winning tickets in 20 tries}] = (1-p)^{20}$ so $P[20]=1-\sum \limits_{k=1}^{19}p(1-p)^{k-1} = p(1-p)^{19} + (1-p)^{20}$ come to think of it the second form is much cleaner so I would write $P_X(k) = \begin{cases}0 &k<1 \\p(1-p)^{k-1} &1\leq k \leq 19\\p(1-p)^{19}+(1-p)^{20} &k=20\\0 &20 < k \end{cases}$
Now, I understand why P[20] cannot equal to p(1-p)^19 because Kathy stop buying tickets when she (a) gets a winner or (b) she has scratched 20 tickets, which ever comes first.
So P[20] = P[bought winning ticket on 20th try] + P[did not buy any winning tickets in 20 tries] = p(1-p)^19 + (1-p)^20 = (1-p)^19.

However, I think this is not correct $P[20]=1-\sum \limits_{k=1}^{19}p(1-p)^{k-1} = p(1-p)^{19} + (1-p)^{20}$ because I calculate each of them and it seems like they are not equal.
Can you take a look again, please. Sorry if I am wrong.

And by the way, can you take a look at my answer for the part II.
P_X(x<17) = P_X(1)+P_X(2)+P_X(3)+.......+P_X(15)+P_X(16)=0.729 7486

Last edited by Shanonhaliwell; March 3rd, 2018 at 04:42 PM.

March 3rd, 2018, 05:50 PM   #6
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 March 3rd, 2018, 07:01 PM #7 Newbie   Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2 Yes, you are right. I have miscalculated the whole thing. My apologies. Thank for your help.

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