Answer:0.075
Step-by-step explanation:
3/8=0.375
0.375/5=0.075
Answer:
![P(X_1,X_2) = \left \{ {{p^2(1-p)^{k+l}, \quad k,l = 0,1,2,\ldots} \atop {0, \quad \text{otherwise}}} \right.](https://tex.z-dn.net/?f=P%28X_1%2CX_2%29%20%3D%20%5Cleft%20%5C%7B%20%7B%7Bp%5E2%281-p%29%5E%7Bk%2Bl%7D%2C%20%5Cquad%20k%2Cl%20%3D%200%2C1%2C2%2C%5Cldots%7D%20%5Catop%20%7B0%2C%20%5Cquad%20%5Ctext%7Botherwise%7D%7D%7D%20%5Cright.)
Step-by-step explanation:
Let
be a random variable that counts the number of failures preceding the first success and
be a random variable that counts the number of failures between the first two successes. The probability of success in a independent trial is
and both of them have Bernoulli distribution.
Assume that the number of unsuccessful trials before the first successful one is
and that
is a number of failures between two successes.
The joint mass function of two discrete random variables
and
is defined by
![P_{XY}(x,y) = P(X=x,Y=y)](https://tex.z-dn.net/?f=P_%7BXY%7D%28x%2Cy%29%20%3D%20P%28X%3Dx%2CY%3Dy%29)
In this case, we have two discrete random variables
and
which are independent, so we have that their joint mass function is of the form
![P_{X_1,X_2}(x,y) = P(X_1=x,X_2=y) = P(X_1=x) \cdot P(X_2=y)](https://tex.z-dn.net/?f=P_%7BX_1%2CX_2%7D%28x%2Cy%29%20%3D%20P%28X_1%3Dx%2CX_2%3Dy%29%20%3D%20P%28X_1%3Dx%29%20%5Ccdot%20P%28X_2%3Dy%29)
The probability that the number of failures before the first successful trial is equal to
is
![P(X_1=k) = \underbrace{(1-p)^k}_{\text{in $k$ trials a failure}} \cdot \underbrace{p}_{\text{after that,a success}}](https://tex.z-dn.net/?f=P%28X_1%3Dk%29%20%3D%20%5Cunderbrace%7B%281-p%29%5Ek%7D_%7B%5Ctext%7Bin%20%24k%24%20trials%20a%20failure%7D%7D%20%5Ccdot%20%5Cunderbrace%7Bp%7D_%7B%5Ctext%7Bafter%20that%2Ca%20success%7D%7D)
The probability that the number of failures between two successes is equal to
is
![P(X_2 = l) = \underbrace{p}_{\text{a success}} \cdot \underbrace{(1-p)^{j}}_{\text{$j$ failures}} = p(1-p)^j](https://tex.z-dn.net/?f=P%28X_2%20%3D%20l%29%20%3D%20%5Cunderbrace%7Bp%7D_%7B%5Ctext%7Ba%20success%7D%7D%20%5Ccdot%20%5Cunderbrace%7B%281-p%29%5E%7Bj%7D%7D_%7B%5Ctext%7B%24j%24%20failures%7D%7D%20%3D%20p%281-p%29%5Ej)
Therefore,
![P(X_1,X_2)= P(X_1=k) \cdot P(X_2=l) \\\phantom{P_{X_1,X_2} \; =} = (1-p)^k \cdot p \cdot p (1-p)^l \\\phantom{P_{X_1,X_2} \; =} = (1-p)^{k+l} \cdot p^2](https://tex.z-dn.net/?f=P%28X_1%2CX_2%29%3D%20%20P%28X_1%3Dk%29%20%5Ccdot%20P%28X_2%3Dl%29%20%5C%5C%5Cphantom%7BP_%7BX_1%2CX_2%7D%20%5C%3B%20%20%3D%7D%20%3D%20%281-p%29%5Ek%20%5Ccdot%20p%20%5Ccdot%20p%20%281-p%29%5El%20%5C%5C%5Cphantom%7BP_%7BX_1%2CX_2%7D%20%5C%3B%20%3D%7D%20%3D%20%281-p%29%5E%7Bk%2Bl%7D%20%5Ccdot%20p%5E2)
Now, we obtain that that their joint mass function is
![P(X_1,X_2) = \left \{ {{p^2(1-p)^{k+l}, \quad k,l = 0,1,2,\ldots} \atop {0, \quad \text{otherwise}}} \right.](https://tex.z-dn.net/?f=P%28X_1%2CX_2%29%20%3D%20%5Cleft%20%5C%7B%20%7B%7Bp%5E2%281-p%29%5E%7Bk%2Bl%7D%2C%20%5Cquad%20k%2Cl%20%3D%200%2C1%2C2%2C%5Cldots%7D%20%5Catop%20%7B0%2C%20%5Cquad%20%5Ctext%7Botherwise%7D%7D%7D%20%5Cright.)
Answer:
9
Step-by-step explanation:
= -3²
= 9
I think it's 11.40
Step-by-step explanation:
hope it's right
196 + x = 500
Subtract 196 from both sides:
196 on the left is cancelled out, and you’re left with x = 304
:)