Assume you sell sandwiches. 70% people choose egg, and the rest choose chicken. Probability of selling 2 egg sandwiches to the next 3 customers?
Clash Royale CLAN TAG#URR8PPP
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A) 0.343
B) 0.063
C) 0.147
D) 0.027
Solution: (C)
"The probability of selling Egg sandwich is 0.7 & that of a chicken sandwich is 0.3. Now, the probability that next 3 customers would order 2 egg sandwich is 0.7 * 0.7 *0.3 = 0.147. They can order them in any sequence, the probabilities would still be the same."
I think the solution is wrong simply because why do we not treat this as a binomial distribution. 3C2 * (0.7) ^2 * (0.3) ? Can someone explain this ?
probability probability-theory
asked Aug 17 at 3:14
Sheldon
161
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up vote
2
down vote
favorite
A) 0.343
B) 0.063
C) 0.147
D) 0.027
Solution: (C)
"The probability of selling Egg sandwich is 0.7 & that of a chicken sandwich is 0.3. Now, the probability that next 3 customers would order 2 egg sandwich is 0.7 * 0.7 *0.3 = 0.147. They can order them in any sequence, the probabilities would still be the same."
I think the solution is wrong simply because why do we not treat this as a binomial distribution. 3C2 * (0.7) ^2 * (0.3) ? Can someone explain this ?
probability probability-theory
asked Aug 17 at 3:14
Sheldon
161
1
I think you're correct. The last sentence of the given solution is true, so far as it goes, but it doesn't explain why you shouldn't add the three probabilities, and indeed, you have to add them.
– saulspatz
Aug 17 at 3:21
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A) 0.343
B) 0.063
C) 0.147
D) 0.027
Solution: (C)
"The probability of selling Egg sandwich is 0.7 & that of a chicken sandwich is 0.3. Now, the probability that next 3 customers would order 2 egg sandwich is 0.7 * 0.7 *0.3 = 0.147. They can order them in any sequence, the probabilities would still be the same."
I think the solution is wrong simply because why do we not treat this as a binomial distribution. 3C2 * (0.7) ^2 * (0.3) ? Can someone explain this ?
probability probability-theory
asked Aug 17 at 3:14
Sheldon
161
A) 0.343
B) 0.063
C) 0.147
D) 0.027
Solution: (C)
"The probability of selling Egg sandwich is 0.7 & that of a chicken sandwich is 0.3. Now, the probability that next 3 customers would order 2 egg sandwich is 0.7 * 0.7 *0.3 = 0.147. They can order them in any sequence, the probabilities would still be the same."
I think the solution is wrong simply because why do we not treat this as a binomial distribution. 3C2 * (0.7) ^2 * (0.3) ? Can someone explain this ?
probability probability-theory
asked Aug 17 at 3:14
Sheldon
161
asked Aug 17 at 3:14
Sheldon
161
asked Aug 17 at 3:14
Sheldon
161
asked Aug 17 at 3:14
Sheldon
161
161
1
I think you're correct. The last sentence of the given solution is true, so far as it goes, but it doesn't explain why you shouldn't add the three probabilities, and indeed, you have to add them.
– saulspatz
Aug 17 at 3:21
add a comment |Â
1
I think you're correct. The last sentence of the given solution is true, so far as it goes, but it doesn't explain why you shouldn't add the three probabilities, and indeed, you have to add them.
– saulspatz
Aug 17 at 3:21
1
1
I think you're correct. The last sentence of the given solution is true, so far as it goes, but it doesn't explain why you shouldn't add the three probabilities, and indeed, you have to add them.
– saulspatz
Aug 17 at 3:21
I think you're correct. The last sentence of the given solution is true, so far as it goes, but it doesn't explain why you shouldn't add the three probabilities, and indeed, you have to add them.
– saulspatz
Aug 17 at 3:21
add a comment |Â
1 Answer
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2
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So assume you have three customers $E,E,C$ .. The way you could see this is by flipping them. So you got three events:
- The first is when they arrive in this order EEC
- The second is ECE
- The third is CEE
So for any of those cases, the probability of any of the above events is
$$P(X = E) times P(X = C) times P(X = E) = p^2 (1-p) = 0.7^2times 0.3$$
Since it happens with 3 combinations above, $C_3^2 = 3$, then
$$C_3^2 p^2 (1-p)$$
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
So assume you have three customers $E,E,C$ .. The way you could see this is by flipping them. So you got three events:
- The first is when they arrive in this order EEC
- The second is ECE
- The third is CEE
So for any of those cases, the probability of any of the above events is
$$P(X = E) times P(X = C) times P(X = E) = p^2 (1-p) = 0.7^2times 0.3$$
Since it happens with 3 combinations above, $C_3^2 = 3$, then
$$C_3^2 p^2 (1-p)$$
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
add a comment |Â
up vote
2
down vote
So assume you have three customers $E,E,C$ .. The way you could see this is by flipping them. So you got three events:
- The first is when they arrive in this order EEC
- The second is ECE
- The third is CEE
So for any of those cases, the probability of any of the above events is
$$P(X = E) times P(X = C) times P(X = E) = p^2 (1-p) = 0.7^2times 0.3$$
Since it happens with 3 combinations above, $C_3^2 = 3$, then
$$C_3^2 p^2 (1-p)$$
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
add a comment |Â
up vote
2
down vote
up vote
2
down vote
So assume you have three customers $E,E,C$ .. The way you could see this is by flipping them. So you got three events:
- The first is when they arrive in this order EEC
- The second is ECE
- The third is CEE
So for any of those cases, the probability of any of the above events is
$$P(X = E) times P(X = C) times P(X = E) = p^2 (1-p) = 0.7^2times 0.3$$
Since it happens with 3 combinations above, $C_3^2 = 3$, then
$$C_3^2 p^2 (1-p)$$
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
So assume you have three customers $E,E,C$ .. The way you could see this is by flipping them. So you got three events:
- The first is when they arrive in this order EEC
- The second is ECE
- The third is CEE
So for any of those cases, the probability of any of the above events is
$$P(X = E) times P(X = C) times P(X = E) = p^2 (1-p) = 0.7^2times 0.3$$
Since it happens with 3 combinations above, $C_3^2 = 3$, then
$$C_3^2 p^2 (1-p)$$
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
answered Aug 17 at 3:20
Ahmad Bazzi
3,1881420
3,1881420
add a comment |Â
add a comment |Â
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1
I think you're correct. The last sentence of the given solution is true, so far as it goes, but it doesn't explain why you shouldn't add the three probabilities, and indeed, you have to add them.
– saulspatz
Aug 17 at 3:21