Find the probability of this event
Two teams play a best-of -seven series, in which the series ends as soon as one team wins four games. The first two games are to be played on A's field , the next three games on B's field, and the last two on A's field. The probability that A wins a game is $0.7$ at home and $0.5$ away. Assume that the games are independent. Find the probability, that the series does not go to $6$ games.
My Answer is
$$2cdot 0.7cdot 0.7cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.3cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\=0.25$$
But the correct answer is $0.397$.
How to get the correct answer?
Thank you very much.
probability statistics
edited Jul 28 at 10:04
Cornman
3,16521229
asked Oct 1 '13 at 2:48
Jing
74211
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
Two teams play a best-of -seven series, in which the series ends as soon as one team wins four games. The first two games are to be played on A's field , the next three games on B's field, and the last two on A's field. The probability that A wins a game is $0.7$ at home and $0.5$ away. Assume that the games are independent. Find the probability, that the series does not go to $6$ games.
My Answer is
$$2cdot 0.7cdot 0.7cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.3cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\=0.25$$
But the correct answer is $0.397$.
How to get the correct answer?
Thank you very much.
probability statistics
edited Jul 28 at 10:04
Cornman
3,16521229
asked Oct 1 '13 at 2:48
Jing
74211
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
2
Did you count for the 2 cases where A or B sweeps the series and wins in 4 games?
– JB King
Oct 1 '13 at 2:53
When asking this question, it would make it easier on the reader if you explain your logic for each term, or at least for the first one. Something like "the chance that A wins two at home and two on the road is ___ given that they have to win the last one for the series to end in five games.
– Ross Millikan
Oct 1 '13 at 2:59
add a comment |
Two teams play a best-of -seven series, in which the series ends as soon as one team wins four games. The first two games are to be played on A's field , the next three games on B's field, and the last two on A's field. The probability that A wins a game is $0.7$ at home and $0.5$ away. Assume that the games are independent. Find the probability, that the series does not go to $6$ games.
My Answer is
$$2cdot 0.7cdot 0.7cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.3cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\=0.25$$
But the correct answer is $0.397$.
How to get the correct answer?
Thank you very much.
probability statistics
edited Jul 28 at 10:04
Cornman
3,16521229
asked Oct 1 '13 at 2:48
Jing
74211
Two teams play a best-of -seven series, in which the series ends as soon as one team wins four games. The first two games are to be played on A's field , the next three games on B's field, and the last two on A's field. The probability that A wins a game is $0.7$ at home and $0.5$ away. Assume that the games are independent. Find the probability, that the series does not go to $6$ games.
My Answer is
$$2cdot 0.7cdot 0.7cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.3cdot 0.3cdot 0.5cdot 0.5cdot 0.5\+2cdot 0.7cdot 0.3cdot 0.5cdot 0.5cdot 0.5\=0.25$$
But the correct answer is $0.397$.
How to get the correct answer?
Thank you very much.
probability statistics
probability statistics
edited Jul 28 at 10:04
Cornman
3,16521229
asked Oct 1 '13 at 2:48
Jing
74211
edited Jul 28 at 10:04
Cornman
3,16521229
asked Oct 1 '13 at 2:48
Jing
74211
edited Jul 28 at 10:04
Cornman
3,16521229
edited Jul 28 at 10:04
Cornman
3,16521229
edited Jul 28 at 10:04
Cornman
3,16521229
3,16521229
asked Oct 1 '13 at 2:48
Jing
74211
asked Oct 1 '13 at 2:48
Jing
74211
asked Oct 1 '13 at 2:48
Jing
74211
74211
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
2
Did you count for the 2 cases where A or B sweeps the series and wins in 4 games?
– JB King
Oct 1 '13 at 2:53
When asking this question, it would make it easier on the reader if you explain your logic for each term, or at least for the first one. Something like "the chance that A wins two at home and two on the road is ___ given that they have to win the last one for the series to end in five games.
– Ross Millikan
Oct 1 '13 at 2:59
add a comment |
2
Did you count for the 2 cases where A or B sweeps the series and wins in 4 games?
– JB King
Oct 1 '13 at 2:53
When asking this question, it would make it easier on the reader if you explain your logic for each term, or at least for the first one. Something like "the chance that A wins two at home and two on the road is ___ given that they have to win the last one for the series to end in five games.
– Ross Millikan
Oct 1 '13 at 2:59
2
2
Did you count for the 2 cases where A or B sweeps the series and wins in 4 games?
– JB King
Oct 1 '13 at 2:53
Did you count for the 2 cases where A or B sweeps the series and wins in 4 games?
– JB King
Oct 1 '13 at 2:53
When asking this question, it would make it easier on the reader if you explain your logic for each term, or at least for the first one. Something like "the chance that A wins two at home and two on the road is ___ given that they have to win the last one for the series to end in five games.
– Ross Millikan
Oct 1 '13 at 2:59
When asking this question, it would make it easier on the reader if you explain your logic for each term, or at least for the first one. Something like "the chance that A wins two at home and two on the road is ___ given that they have to win the last one for the series to end in five games.
– Ross Millikan
Oct 1 '13 at 2:59
add a comment |
1 Answer
1
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votes
We want the probability that the series lasts $4$ or $5$ games.
Team A sweeps in $4$ with probability $(0.7)^2(0.5)^2$, and Team B sweeps with probability $(0.3)^2(0.5)^2$. The probability of a sweep is $0.145$.
The $5$ game analysis is lengthier. Let's first deal with Team A winning in $5$. Team A has to win the fifth game (probability $0.5$, and $3$ of the first $4$. So it has to either (i) win its $2$ home games, and $1$ of the $2$ away games or (ii)
it has to win $1$ home game, and $2$ away games.
Event (i) has probability $(0.7)^2(2)(0.5)^2$. Event (ii) has probability $(2)(0.7)(0.3)(0.5)^2$, for a total of $(0.5)((0.7)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.175$.
A similar calculation for B winning in $5$ gives $(0.5)((0.3)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.075$.
Add up. We get $0.395$.
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
add a comment |
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1 Answer
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We want the probability that the series lasts $4$ or $5$ games.
Team A sweeps in $4$ with probability $(0.7)^2(0.5)^2$, and Team B sweeps with probability $(0.3)^2(0.5)^2$. The probability of a sweep is $0.145$.
The $5$ game analysis is lengthier. Let's first deal with Team A winning in $5$. Team A has to win the fifth game (probability $0.5$, and $3$ of the first $4$. So it has to either (i) win its $2$ home games, and $1$ of the $2$ away games or (ii)
it has to win $1$ home game, and $2$ away games.
Event (i) has probability $(0.7)^2(2)(0.5)^2$. Event (ii) has probability $(2)(0.7)(0.3)(0.5)^2$, for a total of $(0.5)((0.7)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.175$.
A similar calculation for B winning in $5$ gives $(0.5)((0.3)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.075$.
Add up. We get $0.395$.
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
add a comment |
We want the probability that the series lasts $4$ or $5$ games.
Team A sweeps in $4$ with probability $(0.7)^2(0.5)^2$, and Team B sweeps with probability $(0.3)^2(0.5)^2$. The probability of a sweep is $0.145$.
The $5$ game analysis is lengthier. Let's first deal with Team A winning in $5$. Team A has to win the fifth game (probability $0.5$, and $3$ of the first $4$. So it has to either (i) win its $2$ home games, and $1$ of the $2$ away games or (ii)
it has to win $1$ home game, and $2$ away games.
Event (i) has probability $(0.7)^2(2)(0.5)^2$. Event (ii) has probability $(2)(0.7)(0.3)(0.5)^2$, for a total of $(0.5)((0.7)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.175$.
A similar calculation for B winning in $5$ gives $(0.5)((0.3)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.075$.
Add up. We get $0.395$.
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
add a comment |
We want the probability that the series lasts $4$ or $5$ games.
Team A sweeps in $4$ with probability $(0.7)^2(0.5)^2$, and Team B sweeps with probability $(0.3)^2(0.5)^2$. The probability of a sweep is $0.145$.
The $5$ game analysis is lengthier. Let's first deal with Team A winning in $5$. Team A has to win the fifth game (probability $0.5$, and $3$ of the first $4$. So it has to either (i) win its $2$ home games, and $1$ of the $2$ away games or (ii)
it has to win $1$ home game, and $2$ away games.
Event (i) has probability $(0.7)^2(2)(0.5)^2$. Event (ii) has probability $(2)(0.7)(0.3)(0.5)^2$, for a total of $(0.5)((0.7)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.175$.
A similar calculation for B winning in $5$ gives $(0.5)((0.3)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.075$.
Add up. We get $0.395$.
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
We want the probability that the series lasts $4$ or $5$ games.
Team A sweeps in $4$ with probability $(0.7)^2(0.5)^2$, and Team B sweeps with probability $(0.3)^2(0.5)^2$. The probability of a sweep is $0.145$.
The $5$ game analysis is lengthier. Let's first deal with Team A winning in $5$. Team A has to win the fifth game (probability $0.5$, and $3$ of the first $4$. So it has to either (i) win its $2$ home games, and $1$ of the $2$ away games or (ii)
it has to win $1$ home game, and $2$ away games.
Event (i) has probability $(0.7)^2(2)(0.5)^2$. Event (ii) has probability $(2)(0.7)(0.3)(0.5)^2$, for a total of $(0.5)((0.7)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.175$.
A similar calculation for B winning in $5$ gives $(0.5)((0.3)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.075$.
Add up. We get $0.395$.
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
edited Oct 1 '13 at 3:13
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
answered Oct 1 '13 at 3:08
André Nicolas
451k36421805
451k36421805
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
add a comment |
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
You are welcome.
– André Nicolas
Oct 1 '13 at 3:18
add a comment |
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2
Did you count for the 2 cases where A or B sweeps the series and wins in 4 games?
– JB King
Oct 1 '13 at 2:53
When asking this question, it would make it easier on the reader if you explain your logic for each term, or at least for the first one. Something like "the chance that A wins two at home and two on the road is ___ given that they have to win the last one for the series to end in five games.
– Ross Millikan
Oct 1 '13 at 2:59