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Probability high school math competition problem
The Next CEO of Stack OverflowRanking probability taskProbability…why are they different?Can this conditional probability be answered using Bayesian Theorem (or at all) with the information givenProbability Question, choosing a favorable caseExpected Exponential Times: Extended Memoryless Property?Poor Luck in Fantasy FootballConditional Probability going wrongCan someone explain to me why hot hand phenomenon is considered a fallacy?Probability of final outcome of a game of two halvesTwo random sets, how to formulate what percent of set A is bigger than set B.
$begingroup$
Team A and Team B are playing basketball.
Team A starts with the ball, and the ball alternates between the two teams.
When a team has the ball, they have a 50% chance of scoring 1 point.
Regardless of whether or not they score, the ball is given to the other team after they attempt to score.
What is the probability that team A will score 5 points before Team B scores any?
So far I got (1/2)^9 + (1/2)^11 * ((6 choose 1) - 1) + (1/2)^13 * ((7 choose 2) - 1) + ...
I don’t know how to simplify this infinite series or maybe there is better way to solve it
probability probability-theory probability-distributions conditional-probability
asked Mar 18 at 21:58
user653261user653261
411
$endgroup$
add a comment |
$begingroup$
Team A and Team B are playing basketball.
Team A starts with the ball, and the ball alternates between the two teams.
When a team has the ball, they have a 50% chance of scoring 1 point.
Regardless of whether or not they score, the ball is given to the other team after they attempt to score.
What is the probability that team A will score 5 points before Team B scores any?
So far I got (1/2)^9 + (1/2)^11 * ((6 choose 1) - 1) + (1/2)^13 * ((7 choose 2) - 1) + ...
I don’t know how to simplify this infinite series or maybe there is better way to solve it
probability probability-theory probability-distributions conditional-probability
asked Mar 18 at 21:58
user653261user653261
411
$endgroup$
add a comment |
$begingroup$
Team A and Team B are playing basketball.
Team A starts with the ball, and the ball alternates between the two teams.
When a team has the ball, they have a 50% chance of scoring 1 point.
Regardless of whether or not they score, the ball is given to the other team after they attempt to score.
What is the probability that team A will score 5 points before Team B scores any?
So far I got (1/2)^9 + (1/2)^11 * ((6 choose 1) - 1) + (1/2)^13 * ((7 choose 2) - 1) + ...
I don’t know how to simplify this infinite series or maybe there is better way to solve it
probability probability-theory probability-distributions conditional-probability
asked Mar 18 at 21:58
user653261user653261
411
$endgroup$
Team A and Team B are playing basketball.
Team A starts with the ball, and the ball alternates between the two teams.
When a team has the ball, they have a 50% chance of scoring 1 point.
Regardless of whether or not they score, the ball is given to the other team after they attempt to score.
What is the probability that team A will score 5 points before Team B scores any?
So far I got (1/2)^9 + (1/2)^11 * ((6 choose 1) - 1) + (1/2)^13 * ((7 choose 2) - 1) + ...
I don’t know how to simplify this infinite series or maybe there is better way to solve it
probability probability-theory probability-distributions conditional-probability
probability probability-theory probability-distributions conditional-probability
asked Mar 18 at 21:58
user653261user653261
411
asked Mar 18 at 21:58
user653261user653261
411
asked Mar 18 at 21:58
user653261user653261
411
asked Mar 18 at 21:58
user653261user653261
411
asked Mar 18 at 21:58
user653261user653261
411
411
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a nice way to solve this recursively. Let $a_n$ be the probability that team $A$ scores $n$ points before team $B$ scores any. You want $a_5$.
There are two ways this can occur; either $A$ makes the basket, $B$ misses, and then $A$ makes $n-1$ baskets before $B$ makes any, or $A$ misses, $B$ misses, and then $A$ makes $n$ baskets before $B$ makes any. Therefore,
$$
a_n=frac14a_n-1+frac14a_n,tag1
$$
holds for all $nge 2$. However, $a_1$ behaves a little differently; you instead have
$$
a_1=frac12+frac14 a_1tag2
$$
$(2)$ lets you solve for $a_1$, and then you can use that with $(1)$ to get $a_2$, then $a_3$, then $a_4$, then $a_5.$
Here is a more direct method. Initially, there are four possibilities for how the first two shots go:
- Both teams make it.
- A misses, B makes it.
- A makes it, B misses.
- Both teams miss.
If $1$ or $2$ occurs, then team $A$ immediately fails to make $5$ baskets before $B$ makes $1$. If $4$ occurs, then we try again. Since this is repeated until $4$ does not occur, we can ignore $4$; effectively, there are three equally likely options, and we need option $3$.
So, with probability $1/3$, $A$ makes a basket and $B$ misses. Repeat this three more times, and we see there is a $(1/3)^4$ chance that $A$ makes four baskets before $B$ makes any. For the last basket, both options $1$ and $3$ are favorable to $A$, so the probability is $2/3$ for the last basket. Overall, the probability is $(1/3)^4cdot (2/3)$.
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
$endgroup$
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
There is a nice way to solve this recursively. Let $a_n$ be the probability that team $A$ scores $n$ points before team $B$ scores any. You want $a_5$.
There are two ways this can occur; either $A$ makes the basket, $B$ misses, and then $A$ makes $n-1$ baskets before $B$ makes any, or $A$ misses, $B$ misses, and then $A$ makes $n$ baskets before $B$ makes any. Therefore,
$$
a_n=frac14a_n-1+frac14a_n,tag1
$$
holds for all $nge 2$. However, $a_1$ behaves a little differently; you instead have
$$
a_1=frac12+frac14 a_1tag2
$$
$(2)$ lets you solve for $a_1$, and then you can use that with $(1)$ to get $a_2$, then $a_3$, then $a_4$, then $a_5.$
Here is a more direct method. Initially, there are four possibilities for how the first two shots go:
- Both teams make it.
- A misses, B makes it.
- A makes it, B misses.
- Both teams miss.
If $1$ or $2$ occurs, then team $A$ immediately fails to make $5$ baskets before $B$ makes $1$. If $4$ occurs, then we try again. Since this is repeated until $4$ does not occur, we can ignore $4$; effectively, there are three equally likely options, and we need option $3$.
So, with probability $1/3$, $A$ makes a basket and $B$ misses. Repeat this three more times, and we see there is a $(1/3)^4$ chance that $A$ makes four baskets before $B$ makes any. For the last basket, both options $1$ and $3$ are favorable to $A$, so the probability is $2/3$ for the last basket. Overall, the probability is $(1/3)^4cdot (2/3)$.
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
$endgroup$
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
add a comment |
$begingroup$
There is a nice way to solve this recursively. Let $a_n$ be the probability that team $A$ scores $n$ points before team $B$ scores any. You want $a_5$.
There are two ways this can occur; either $A$ makes the basket, $B$ misses, and then $A$ makes $n-1$ baskets before $B$ makes any, or $A$ misses, $B$ misses, and then $A$ makes $n$ baskets before $B$ makes any. Therefore,
$$
a_n=frac14a_n-1+frac14a_n,tag1
$$
holds for all $nge 2$. However, $a_1$ behaves a little differently; you instead have
$$
a_1=frac12+frac14 a_1tag2
$$
$(2)$ lets you solve for $a_1$, and then you can use that with $(1)$ to get $a_2$, then $a_3$, then $a_4$, then $a_5.$
Here is a more direct method. Initially, there are four possibilities for how the first two shots go:
- Both teams make it.
- A misses, B makes it.
- A makes it, B misses.
- Both teams miss.
If $1$ or $2$ occurs, then team $A$ immediately fails to make $5$ baskets before $B$ makes $1$. If $4$ occurs, then we try again. Since this is repeated until $4$ does not occur, we can ignore $4$; effectively, there are three equally likely options, and we need option $3$.
So, with probability $1/3$, $A$ makes a basket and $B$ misses. Repeat this three more times, and we see there is a $(1/3)^4$ chance that $A$ makes four baskets before $B$ makes any. For the last basket, both options $1$ and $3$ are favorable to $A$, so the probability is $2/3$ for the last basket. Overall, the probability is $(1/3)^4cdot (2/3)$.
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
$endgroup$
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
add a comment |
$begingroup$
There is a nice way to solve this recursively. Let $a_n$ be the probability that team $A$ scores $n$ points before team $B$ scores any. You want $a_5$.
There are two ways this can occur; either $A$ makes the basket, $B$ misses, and then $A$ makes $n-1$ baskets before $B$ makes any, or $A$ misses, $B$ misses, and then $A$ makes $n$ baskets before $B$ makes any. Therefore,
$$
a_n=frac14a_n-1+frac14a_n,tag1
$$
holds for all $nge 2$. However, $a_1$ behaves a little differently; you instead have
$$
a_1=frac12+frac14 a_1tag2
$$
$(2)$ lets you solve for $a_1$, and then you can use that with $(1)$ to get $a_2$, then $a_3$, then $a_4$, then $a_5.$
Here is a more direct method. Initially, there are four possibilities for how the first two shots go:
- Both teams make it.
- A misses, B makes it.
- A makes it, B misses.
- Both teams miss.
If $1$ or $2$ occurs, then team $A$ immediately fails to make $5$ baskets before $B$ makes $1$. If $4$ occurs, then we try again. Since this is repeated until $4$ does not occur, we can ignore $4$; effectively, there are three equally likely options, and we need option $3$.
So, with probability $1/3$, $A$ makes a basket and $B$ misses. Repeat this three more times, and we see there is a $(1/3)^4$ chance that $A$ makes four baskets before $B$ makes any. For the last basket, both options $1$ and $3$ are favorable to $A$, so the probability is $2/3$ for the last basket. Overall, the probability is $(1/3)^4cdot (2/3)$.
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
$endgroup$
There is a nice way to solve this recursively. Let $a_n$ be the probability that team $A$ scores $n$ points before team $B$ scores any. You want $a_5$.
There are two ways this can occur; either $A$ makes the basket, $B$ misses, and then $A$ makes $n-1$ baskets before $B$ makes any, or $A$ misses, $B$ misses, and then $A$ makes $n$ baskets before $B$ makes any. Therefore,
$$
a_n=frac14a_n-1+frac14a_n,tag1
$$
holds for all $nge 2$. However, $a_1$ behaves a little differently; you instead have
$$
a_1=frac12+frac14 a_1tag2
$$
$(2)$ lets you solve for $a_1$, and then you can use that with $(1)$ to get $a_2$, then $a_3$, then $a_4$, then $a_5.$
Here is a more direct method. Initially, there are four possibilities for how the first two shots go:
- Both teams make it.
- A misses, B makes it.
- A makes it, B misses.
- Both teams miss.
If $1$ or $2$ occurs, then team $A$ immediately fails to make $5$ baskets before $B$ makes $1$. If $4$ occurs, then we try again. Since this is repeated until $4$ does not occur, we can ignore $4$; effectively, there are three equally likely options, and we need option $3$.
So, with probability $1/3$, $A$ makes a basket and $B$ misses. Repeat this three more times, and we see there is a $(1/3)^4$ chance that $A$ makes four baskets before $B$ makes any. For the last basket, both options $1$ and $3$ are favorable to $A$, so the probability is $2/3$ for the last basket. Overall, the probability is $(1/3)^4cdot (2/3)$.
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
edited Mar 18 at 22:40
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
answered Mar 18 at 22:05
Mike EarnestMike Earnest
26.2k22151
26.2k22151
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
add a comment |
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
$begingroup$
Thx a lot for both answers especially the direct method. Great solution since easy to understand
$endgroup$
– user653261
Mar 19 at 22:33
add a comment |
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