Prize Probability
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The question is as follows:
"Each meal at a fast food restaurant comes with a prize. There are three types of prizes, $A$, $B$, and $C$. Each meal comes with prize $A$ with probability $0.5$, prize $B$ with probability $0.4$, and prize $C$ with probability $0.1$, independently of previous meals.
You buy four meals. What’s the probability that you get a prize of each type?"
This is a very basic question, but I can't figure it out. My attempt was to do $4$ choose $3$ multiplied by each of the probabilites, $0.1, 0.4, 0.5$, but $0.8$ is incorrect. Not sure how to approach this problem. Intution behind solution would be appreciated
probability combinatorics
asked Nov 27 '18 at 23:10
reeree
216
$endgroup$
add a comment |
$begingroup$
The question is as follows:
"Each meal at a fast food restaurant comes with a prize. There are three types of prizes, $A$, $B$, and $C$. Each meal comes with prize $A$ with probability $0.5$, prize $B$ with probability $0.4$, and prize $C$ with probability $0.1$, independently of previous meals.
You buy four meals. What’s the probability that you get a prize of each type?"
This is a very basic question, but I can't figure it out. My attempt was to do $4$ choose $3$ multiplied by each of the probabilites, $0.1, 0.4, 0.5$, but $0.8$ is incorrect. Not sure how to approach this problem. Intution behind solution would be appreciated
probability combinatorics
asked Nov 27 '18 at 23:10
reeree
216
$endgroup$
$begingroup$
For the favorable event to happen, you must win one prize twice and each of the others once. Are you familiar with the multinomial distribution?
$endgroup$
– N. F. Taussig
Nov 27 '18 at 23:16
add a comment |
$begingroup$
The question is as follows:
"Each meal at a fast food restaurant comes with a prize. There are three types of prizes, $A$, $B$, and $C$. Each meal comes with prize $A$ with probability $0.5$, prize $B$ with probability $0.4$, and prize $C$ with probability $0.1$, independently of previous meals.
You buy four meals. What’s the probability that you get a prize of each type?"
This is a very basic question, but I can't figure it out. My attempt was to do $4$ choose $3$ multiplied by each of the probabilites, $0.1, 0.4, 0.5$, but $0.8$ is incorrect. Not sure how to approach this problem. Intution behind solution would be appreciated
probability combinatorics
asked Nov 27 '18 at 23:10
reeree
216
$endgroup$
The question is as follows:
"Each meal at a fast food restaurant comes with a prize. There are three types of prizes, $A$, $B$, and $C$. Each meal comes with prize $A$ with probability $0.5$, prize $B$ with probability $0.4$, and prize $C$ with probability $0.1$, independently of previous meals.
You buy four meals. What’s the probability that you get a prize of each type?"
This is a very basic question, but I can't figure it out. My attempt was to do $4$ choose $3$ multiplied by each of the probabilites, $0.1, 0.4, 0.5$, but $0.8$ is incorrect. Not sure how to approach this problem. Intution behind solution would be appreciated
probability combinatorics
probability combinatorics
asked Nov 27 '18 at 23:10
reeree
216
asked Nov 27 '18 at 23:10
reeree
216
asked Nov 27 '18 at 23:10
reeree
216
asked Nov 27 '18 at 23:10
reeree
216
asked Nov 27 '18 at 23:10
reeree
216
216
$begingroup$
For the favorable event to happen, you must win one prize twice and each of the others once. Are you familiar with the multinomial distribution?
$endgroup$
– N. F. Taussig
Nov 27 '18 at 23:16
add a comment |
$begingroup$
For the favorable event to happen, you must win one prize twice and each of the others once. Are you familiar with the multinomial distribution?
$endgroup$
– N. F. Taussig
Nov 27 '18 at 23:16
$begingroup$
For the favorable event to happen, you must win one prize twice and each of the others once. Are you familiar with the multinomial distribution?
$endgroup$
– N. F. Taussig
Nov 27 '18 at 23:16
$begingroup$
For the favorable event to happen, you must win one prize twice and each of the others once. Are you familiar with the multinomial distribution?
$endgroup$
– N. F. Taussig
Nov 27 '18 at 23:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to account for the fact that, if you get all three prizes, one of them comes twice. So you could get $AABC$, $CBCA$, etc. The number of ways to choose a permutation of two $A$'s, one $B$, and a $C$ is $binom{4}{2} times 2 times 1 = 12$ (the number of ways to count the other similar permutations is the same). The probability of any such permutation occurring is $(0.5)^2(0.4)(0.1) = 0.01$. Similarly, the probability of a permutation where $B$ is repeated is $(0.5)(0.4)^2(0.1) = 0.008$ and the probability of a permutation where $C$ is repeated is $(0.5)(0.4)(0.1)^2 = 0.002$. Adding over all such permutations gives a $0.24$ probability of getting all three prizes.
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
$endgroup$
add a comment |
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$begingroup$
You need to account for the fact that, if you get all three prizes, one of them comes twice. So you could get $AABC$, $CBCA$, etc. The number of ways to choose a permutation of two $A$'s, one $B$, and a $C$ is $binom{4}{2} times 2 times 1 = 12$ (the number of ways to count the other similar permutations is the same). The probability of any such permutation occurring is $(0.5)^2(0.4)(0.1) = 0.01$. Similarly, the probability of a permutation where $B$ is repeated is $(0.5)(0.4)^2(0.1) = 0.008$ and the probability of a permutation where $C$ is repeated is $(0.5)(0.4)(0.1)^2 = 0.002$. Adding over all such permutations gives a $0.24$ probability of getting all three prizes.
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
You need to account for the fact that, if you get all three prizes, one of them comes twice. So you could get $AABC$, $CBCA$, etc. The number of ways to choose a permutation of two $A$'s, one $B$, and a $C$ is $binom{4}{2} times 2 times 1 = 12$ (the number of ways to count the other similar permutations is the same). The probability of any such permutation occurring is $(0.5)^2(0.4)(0.1) = 0.01$. Similarly, the probability of a permutation where $B$ is repeated is $(0.5)(0.4)^2(0.1) = 0.008$ and the probability of a permutation where $C$ is repeated is $(0.5)(0.4)(0.1)^2 = 0.002$. Adding over all such permutations gives a $0.24$ probability of getting all three prizes.
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
You need to account for the fact that, if you get all three prizes, one of them comes twice. So you could get $AABC$, $CBCA$, etc. The number of ways to choose a permutation of two $A$'s, one $B$, and a $C$ is $binom{4}{2} times 2 times 1 = 12$ (the number of ways to count the other similar permutations is the same). The probability of any such permutation occurring is $(0.5)^2(0.4)(0.1) = 0.01$. Similarly, the probability of a permutation where $B$ is repeated is $(0.5)(0.4)^2(0.1) = 0.008$ and the probability of a permutation where $C$ is repeated is $(0.5)(0.4)(0.1)^2 = 0.002$. Adding over all such permutations gives a $0.24$ probability of getting all three prizes.
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
$endgroup$
You need to account for the fact that, if you get all three prizes, one of them comes twice. So you could get $AABC$, $CBCA$, etc. The number of ways to choose a permutation of two $A$'s, one $B$, and a $C$ is $binom{4}{2} times 2 times 1 = 12$ (the number of ways to count the other similar permutations is the same). The probability of any such permutation occurring is $(0.5)^2(0.4)(0.1) = 0.01$. Similarly, the probability of a permutation where $B$ is repeated is $(0.5)(0.4)^2(0.1) = 0.008$ and the probability of a permutation where $C$ is repeated is $(0.5)(0.4)(0.1)^2 = 0.002$. Adding over all such permutations gives a $0.24$ probability of getting all three prizes.
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
edited Nov 27 '18 at 23:23
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
answered Nov 27 '18 at 23:17
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
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$begingroup$
For the favorable event to happen, you must win one prize twice and each of the others once. Are you familiar with the multinomial distribution?
$endgroup$
– N. F. Taussig
Nov 27 '18 at 23:16