2
$\begingroup$

Suppose two basketball players throw alternately to a basket (infinity times). Player A has probability of 0.7 to score, player B has probability of 0.4 to score. Player A is starting. What is the probability that the 2 first scores were together?

$\endgroup$
4
  • $\begingroup$ What do you mean "were together"? Do you mean they have the same number of hits after 2 attpemts? $\endgroup$
    – Stefan4024
    Dec 6 '13 at 13:01
  • $\begingroup$ if player A did his first score on the n trial, it means that player B did his first score on the n trial too. $\endgroup$ Dec 6 '13 at 13:03
  • $\begingroup$ @AviadChmelnik Are you certain? For instance, if player A misses, player B scores, and then player A scores, I would count that as 'together' even though they scored on different shot-numbers. $\endgroup$ Dec 12 '13 at 21:25
  • $\begingroup$ it looks like you are right steven! $\endgroup$ Dec 13 '13 at 13:17
1
$\begingroup$

In order Player A to score for first time on the $n^{th}$ attempt, he needs to miss the first $n-1$ attempts. The probabilty for miss is $\frac 3{10}$ and for a hit it's $\frac 7{10}$. So the probablity to make his firs basket after $n$ attempts is:

$$P_A(n) = \left(\frac{3}{10}\right)^{n-1}\frac7{10}$$

Simularly for Player B we have:

$$P_B(n) = \left(\frac{3}{5}\right)^{n-1}\frac2{5}$$

If you want to get the probabilty that they'll hit their first shots at specific attempt $n$ is:

$$P_A(n) \cdot P_B(n) = \left(\frac{9}{50}\right)^{n-1}\frac{14}{50}$$

If you want to know the probability of both players making the first shot at any attepmt then the probability is expresed with:

$$\left(\frac{9}{50}\right)^{1-1}\frac{14}{50} + \left(\frac{9}{50}\right)^{2-1}\frac{14}{50} + \left(\frac{9}{50}\right)^{3-1}\frac{14}{50} + \cdots$$

This is simple geometric series and finding the sum as $n \to \infty$ won't be a problem, right?

$\endgroup$
0
$\begingroup$

The chance that both score at very first attempt = 0.7*0.4

at second attempt = (0*3*0.6)*(0.7*0.4) (Both have to fail their first attemts and succeed in second attempt)

at nth attempt is (0.3*0.6)^n *(0.7*0.4) (Both have to fail there n attempts and suceed at n+1 th attempt)

The combined probabillity hence is 0.28 [1+0.18 + 0.18^2...] ~ 0.341

$\endgroup$
1
  • $\begingroup$ Welcome to math.se! You may improve your formatting using mathjax. $\endgroup$ Dec 6 '13 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.