Respuesta :
Answer:
P(X + Y ≤ 2) = 0.387
Step-by-step explanation:
We have to calculate the probability of a sum of random variables.
In this case, the number of combinations is small, so we can do it manually.
The probability of having k mistakes for Anne can be modeled as:
[tex]P(x=k)=\frac{\lambda e^{-\lambda}}{k!}[/tex]
And the probability of having k mistakes for John is*:
[tex]P(y=k)=(1-p)^{k-1}p[/tex]
*This involves assuming that with probability one John makes at least one mistake, P(X≥1)=1.
Then we have to compute all the combinations for X+Y≤2
- X=0; Y=1
- X=1; Y=1
- X=0; Y=2
Now we can calculate:
[tex]P(X + Y \leq2)=P(X=0\&Y=1)+P(X=1\&Y=1)+P(X=0\&Y=2)\\\\\\P(X + Y \leq2)=\frac{\lambda^0 e^{-\lambda}}{0!}*(1-p)^{1-1}p+\frac{\lambda^1 e^{-\lambda}}{1!}*(1-p)^{1-1}p+\frac{\lambda^0 e^{-\lambda}}{0!}*(1-p)^{2-1}p\\\\\\P(X + Y \leq2)=\frac{1^0 e^{-1}}{0!}*(1-0.4)^{0}*0.4+\frac{1^1 e^{-1}}{1!}*(1-0.4)^{0}*0.4+\frac{1^0 e^{-1}}{0!}*(1-0.4)*0.4\\\\\\P(X + Y \leq2)=0.368*0.4+0.368*0.4+0.368*0.24\\\\\\P(X + Y \leq2)=0.147+0.147+0.088=0.387[/tex]
