1
$\begingroup$

Let $X_1, X_2, X_3, X_4$ be independent Bernoulli random variables. Then \begin{align} Pr[X_i=1]=Pr[X_i=0]=1/2. \end{align} I want to compute the following probability \begin{align} Pr( X_1+X_2+X_3=2, X_2+X_4=1 ). \end{align} My solution: Suppose that $X_1+X_2+X_3=2$ and $X_2+X_4=1$. Then $(X_2, X_4)=(0,1)$ or $(1,0)$. The probabilities of $(X_2, X_4)=(0,1)$ and $(1,0)$ are both $1/2 \times 1/2 = 1/4$. If $(X_2, X_4)=(0,1)$, then $X_1+X_2+X_3=X_1+X_3=2$. Therefore $X_1 = X_3 =1$. This happens with probability $1/4$. If $(X_2, X_4)=(1,0)$, then $X_1+X_2+X_3=X_1+1+X_3=2$. Therefore $(X_1, X_3) \in \{(1,0), (0,1)\}$. This happens with probability $1/4$. Therefore \begin{align} Pr( X_1+X_2+X_3=2, X_2+X_4=1 ) = 1/16+2/16=3/16. \end{align} Is this correct? Thank you very much.

$\endgroup$
2
$\begingroup$

Let's verify your result by simulation using a Python script:

import numpy as np

N = 10**5 # number of trials

# list of N 4-ples (X1, X2, X3, X4)
XX = [np.random.randint(2, size=4) for n in np.arange(N)]

# list of trials outcomes
P = list(map(lambda X: (X[0]+X[1]+X[2]==2)&(X[2]+X[3]==1), XX))

# average of successes 
np.mean(P) # ~ 0.18772

Since $\frac{3}{16} \simeq 0.18772$ this confirms your result.

$\endgroup$
  • $\begingroup$ thank you very much. Your codes are very helpful. Are these python codes? $\endgroup$ – LJR Aug 4 '18 at 13:20
  • $\begingroup$ @LJR You're welcome! Yes they are! I'll edit the answer $\endgroup$ – Giulio Scattolin Aug 4 '18 at 13:21
2
$\begingroup$

Just add as a supplementary technique:

You may also try to use the law of total probability with conditioning on $X_2$, since only $X_2$ are in common of the two events, then make use of the independence:

$$ \begin{align} &\Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1\} \\ =& \Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1|X_2 = 0\}\Pr\{X_2 = 0\} \\ &+ \Pr\{X_1 + X_2 + X_3 = 2, X_2 + X_4 = 1|X_2 = 1\}\Pr\{X_2 = 1\} \\ =& \Pr\{X_1 + X_3 = 2, X_4 = 1|X_2 = 0\}\Pr\{X_2 = 0\} \\ &+ \Pr\{X_1 + X_3 = 1, X_4 = 0|X_2 = 1\}\Pr\{X_2 = 1\} \\ =& \Pr\{X_1 + X_3 = 2\}\Pr\{X_4 = 1\}\Pr\{X_2 = 0\} + \\ &+ \Pr\{X_1 + X_3 = 1\}\Pr\{X_4 = 0\}\Pr\{X_2 = 1\} \\ =& \frac {1} {4}\times \frac {1} {2} \times \frac {1} {2} + \frac {1} {2} \times \frac {1} {2} \times \frac {1} {2} \\ =& \frac {3} {16} \end{align}$$

So essentially you have counted the cases, just like what you have did.

$\endgroup$
1
$\begingroup$

Yes, your solution is indeed correct!


Minor comment: In the last sentence before the final expression "this" is probably referred to each of the outcomes $(1, 0)$ and $(0, 1)$ rather than to the event $(X_1, X_3) \in \{ (1, 0), (0, 1) \}$.

$\endgroup$
  • 1
    $\begingroup$ @Pointguardo, yes, thank you very much for your comments. $\endgroup$ – LJR Aug 4 '18 at 12:20

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.