Find probability using geometric distributionTwo boys play basketball in the following way. They take turns shooting and stop when a basket is made. Player A goes first and has probability p 1 of making a basket on any throw. Player B, who shoots second, has probability p 2 of making a basket. The outcomes of the successive trials are assumed to be independent.a. Find the frequency function for the total number of attempts.b. What is the probability that player A wins?My solution: a. suppose that they make k attempts. Then the answer depends on parity of k. If k = 2 n + 1 it means that first player made n + 1 attempts and n first attempts were unsuccessful, while the last one was successful (he made a basket). As for second player, he made n unsuccessful attempts. Bot these probabilities are easily found by multiplication of respective probabilities for the first and second player.The case when k is even is solved in a similar way.b. According to part a., we know how to find the probability that the game is over at 2 n + 1 attempt. Therefore we sum these probabilities over n ∈ N .Is it correct?

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Find probability using geometric distributionTwo boys play basketball in the following way. They take turns shooting and stop when a basket is made. Player A goes first and has probability       p    1   of making a basket on any throw. Player B, who shoots second, has probability       p    2   of making a basket. The outcomes of the successive trials are assumed to be independent.a. Find the frequency function for the total number of attempts.b. What is the probability that player A wins?My solution: a. suppose that they make k attempts. Then the answer depends on parity of k. If   k  =  2  n  +  1 it means that first player made   n  +  1 attempts and n first attempts were unsuccessful, while the last one was successful (he made a basket). As for second player, he made n unsuccessful attempts. Bot these probabilities are easily found by multiplication of respective probabilities for the first and second player.The case when k is even is solved in a similar way.b. According to part a., we know how to find the probability that the game is over at   2  n  +  1 attempt. Therefore we sum these probabilities over   n  ∈      N  .Is it correct?

bigfreakystargl · Accepted Answer

Step 1Nothing wrong with enumerating all the paths and then summing the probabilities. Another approach, which is sometimes simpler algebraically, is to do it recursively:Let   P  (      p    1    ,      p    2    ) be the probability A wins given the data you mention. Let&#039;s say A takes one shot. It goes in, with probability       p    1   and misses with probability   1  −      p    1   If it misses, you are in back at the start of a similar game...only now Player B shoots first and the probability that A wins from here is 1 -   P  (      p    2    ,      p    1    ). Thus   P  (      p    1    ,      p    2    )  =      p    1    +  (  1  −      p    1    )  (  1  −  P  (      p    2    ,      p    1    )  )Step 2A similar calculation shows that   P  (      p    2    ,      p    1    )  =      p    2    +  (  1  −      p    2    )  (  1  −  P  (      p    1    ,      p    2    )  )In this way we get two equations in two unknowns which can easily be solved to yield   P  (      p    1    ,      p    2    )  =            p      1                      p        1            −              p        1                    p        2            +              p        2            (trusting that no algebraic error was made). As a sanity check, if we assume both probabilities are       1    2  , as in the case of alternating coin tosses, then this becomes       2    3   which is the familiar answer.

Madilyn Quinn · Accepted Answer

Step 1Odd number of attempts ending with success results that the first player wins with probability       p    1        ∑          n      =      0              ∞        [  (  1  −      p    1    )  (  1  −      p    2    )      ]    n    =            p      1                      p        1            +              p        2            −              p        1                    p        2              ..So, the probability that the second player wins is 1 minus the probability above:                     p        2            (      1      −              p        1            )                      p        1            +              p        2            −              p        1                    p        2              ..Step 2Or, one can say that if the number of attempts ending with a success is even then the second player wins. The probability of this event is       p    2        ∑          n      =      1              ∞        (  1  −      p    1        )    n    (  1  −      p    2        )          n      −      1        =      p    2    (  1  −      p    1    )      ∑          n      =      0              ∞        [  (  1  −      p    1    )  (  1  −      p    2    )      ]    n    =                    p        2            (      1      −              p        1            )                      p        1            +              p        2            −              p        1                    p        2              .

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