Free throws in basketball game - Probability The random variable $X_i$ indicates whether in a basketball game the $i$-th free throw is hit ($X_i = 1$) or not ($X_i = 0$).
$10$ free throws are executed, so that $i = 1,\ldots 10$.
It holds that $P(X_i=0)=0,3$ and $P(X_i=1)=0,7$. 
X denotes the sum of these ten independent random variables: $X = \sum_{i=1}^{10}X_i$. 
(a) Determine $P (X = 8)$. What does this probability mean in terms of content? 
(b) What is the probability that the number of hits is at most five? 
(c) Which number of hits can we expect on average? 
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(a) Does it hold that $P(X=8)=0,3^2\cdot 0,7^8=0.0051883209$ ? That would mean that the probability that 8 of 10 free throws were successful is equal to $0,5\%$. 
Is this correct? 
(b) We want to calculate the probability $P(X\geq 5)$, which is equal to $P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=0,3^9\cdot 0,7^1+0,3^8\cdot 0,7^2+0,3^7\cdot 0,7^3+0,3^6\cdot 0,7^4+0,3^5\cdot 0,7^5=0.0007043841$, or not? 
(c) We are asked to calculate the expected value, aren't we? Is this equal to $$E(X)=\sum_{i=1}^{10}x_i\cdot P(X=x_i)\\ =1\cdot P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)+\ldots 9\cdot P(X=9)+10\cdot P(X=10)$$ or not? 
 A: The key for the whole question is to identify the distribution of $X$ which is a sum of i.i.d of Bernoulli distribution. 
Hence $X \sim Bin(10, 0.7)$.
With this tool, you should be able to solve every part.
In particular
$$Pr(X=x) = \binom{10}x 0.7^x\cdot 0.3^{10-x}$$
also, we know the closed form formula for its expectation.
A: (a) Your calculation of P(X = 8) is erroneous. You failed to consider the fact that $0,3^2⋅0,7^8$ only considers 1 case of 8 hits and 2 misses. e.g. HHHHHHHHMM. In fact, there are $\frac{10!}{2!8!}$ cases (using combinatorics), leading to $\frac{10!}{2!8!}*(0,3^2⋅0,7^8)$ being the correct solution for P(X=8). Your definition of the probability is also a bit vague. I would phrase it in this way "the probability that 8 of 10 free throws were successfully hit"
(b) You have misinterpreted the question. The number of hits is at most 5, so X should be at most 5. Hence, we should find the $P(X\le 5)$. You would have to use a Binomial Distribution for this. There are 10 independent free throws, with a P(successful hit) = $P(X_i=1)$ = 0.7. Hence, X~B(10,0.7). You can use a calculator or Excel to find $P(X\le 5)$ from there. (BinomCDF)
(c) E(X) = np for a Binomial Distribution. In this case, E(X) = 10 * 0.7 = 7. For a more comprehensive answer, we can use the general formula E(X) = $\sum_{i=1}^{10} x_ip_i$. Find all the $P(X=x_i)$ and mutiply it with all the corresponding $x_i$ values. Sum the product up to find E(X).
Hope this helps.
A: First of all, you must be able to determine that $X\sim Binomial (n=10,p=0.7)$, where $p$ is the probability of a made free throw. From there, you can proceed by noting that $P[X=x]= {n\choose k}(1-p)^{n-k}p^k={10\choose k}(0.3)^{10-k}(0.7)^k$. You can also interpret $P[X=x]$ as the probability that the number of made free throws out of 10 is $x$.
Then
(a) Calculate $P[X=8]$. Do the interpretation. :)
(b) Calculate $P[X\le 5]=P[X=0]+P[X=1]+P[X=2]+P[X=3]+P[X=4]+P[X=5]$.
(c) The mean of a $X\sim Binomial(n,p)$ is $np$.
A: For the first part, $0.3^2.0.7^8$ represents the probability that 8 out of 10 free throws are made, but you can arrange $8$ made FT and $2$ missed FT in different ways, specifically $^{10}C_2$ ways, so the actual probability is $^{10}C_2.0.3^2.0.7^8$.
In other words, $0.3^2.0.7^8$ represents the probability of hitting $8$ out of $10$ FT in just one of the many possible ways (e.g. hitting the first $8$ FT and missing the last $2$).
Similarly you can solve the second part, but note that you have to calculate $P(X \leq 5)$ (at most $5$ FT).
For the third part, use linearity of expectation - $E[X] = \sum_iE[X_i]$. Alternatively, as already noted by you and Producer of BS, the experiment can be described by a binomial probability distribution, whose expectation is $E[X = n] = np$, where $p$ is the probability of a made FT.
