Probabilities are used to determine the chances of events.
The given parameters:
---- the number of candy bars
---- the probability of selling a candy bar
<u>(a) Name distributions</u>
The distribution of X is represented as:
![X \sim(r,p)](https://tex.z-dn.net/?f=X%20%5Csim%28r%2Cp%29)
Where:
![r = 5](https://tex.z-dn.net/?f=r%20%3D%205)
![p= \frac{5\times 40\%}{10}](https://tex.z-dn.net/?f=p%3D%20%5Cfrac%7B5%5Ctimes%2040%5C%25%7D%7B10%7D)
![p= 0.2](https://tex.z-dn.net/?f=p%3D%200.2)
So, the name distribution of X is ![X \sim(r= 5,p = 0.2)](https://tex.z-dn.net/?f=X%20%5Csim%28r%3D%205%2Cp%20%3D%200.2%29)
<u>(b) The probability that the last candy is sold at the 11th house</u>
This means that:
--- the number of previous houses
--- the previous number of candies
--- the given probability of selling a candy
The probability is calculated using:
![P(x = n+1) = ^{n}C_r \times p^{r +1} \times (1 - p)^{n-r}](https://tex.z-dn.net/?f=P%28x%20%3D%20n%2B1%29%20%3D%20%5E%7Bn%7DC_r%20%5Ctimes%20p%5E%7Br%20%2B1%7D%20%5Ctimes%20%281%20-%20p%29%5E%7Bn-r%7D)
This gives
![P(x = 10+1) = ^{10}C_4 \times 0.4^{4 +1} \times (1 - 0.4)^{10-4}](https://tex.z-dn.net/?f=P%28x%20%3D%2010%2B1%29%20%3D%20%5E%7B10%7DC_4%20%5Ctimes%200.4%5E%7B4%20%2B1%7D%20%5Ctimes%20%281%20-%200.4%29%5E%7B10-4%7D)
![P(x = 11) = ^{10}C_4 \times 0.4^{5} \times (0.6)^6](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%20%5E%7B10%7DC_4%20%5Ctimes%200.4%5E%7B5%7D%20%5Ctimes%20%280.6%29%5E6)
![P(x = 11) = 210 \times 0.4^5 \times 0.6^6](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%20210%20%5Ctimes%200.4%5E5%20%5Ctimes%200.6%5E6)
![P(x = 11) = 0.1003290624](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%200.1003290624)
Approximate
![P(x = 11) = 0.1003](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%200.1003)
Hence, the probability that the last candy is sold at the 11th house is 0.1003
<u>(b) The probability he sells the candies on or before the 8th house</u>
The probability is calculated using:
![P(x \le 8) = P(5 \le x \le 8)](https://tex.z-dn.net/?f=P%28x%20%5Cle%208%29%20%3D%20P%285%20%5Cle%20x%20%5Cle%208%29)
This gives
![P(x \le 8) = ^{10}C_5 \times 0.4^{6} \times (0.6)^5 + ^{10}C_6 \times 0.4^{7} \times (0.6)^4 + ^{10}C_7 \times 0.4^{8} \times (0.6)^3 +^{10}C_8 \times 0.4^{9} \times (0.6)^2](https://tex.z-dn.net/?f=P%28x%20%5Cle%208%29%20%3D%20%5E%7B10%7DC_5%20%5Ctimes%200.4%5E%7B6%7D%20%5Ctimes%20%280.6%29%5E5%20%2B%20%5E%7B10%7DC_6%20%5Ctimes%200.4%5E%7B7%7D%20%5Ctimes%20%280.6%29%5E4%20%2B%20%5E%7B10%7DC_7%20%5Ctimes%200.4%5E%7B8%7D%20%5Ctimes%20%280.6%29%5E3%20%2B%5E%7B10%7DC_8%20%5Ctimes%200.4%5E%7B9%7D%20%5Ctimes%20%280.6%29%5E2)
---- approximated
Hence, the probability he sells the candies on or before the 8th house is 0.1737
Read more about probabilities at:
brainly.com/question/251701