Ch 3 - 13, 17, 31

3.13

Question

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Pasted image 20241003170300.png

Proof
a.

\begin{aligned} P (X \leq 3) & = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) \\ = .10 + .15 + .20 + .25 \\ = .70 \end{aligned}

P (X < 3) = .10 + .15 + .20 = .45

P (X \geq 3) = .25 + .20 + .06 + .04 = .55

P (2 \leq X \leq 5) = .20 + .25 + .20 + .06 = .71

\begin{aligned} P (2 \leq X \leq 4) & = P (X = 2) + P (X = 3) + P (X = 4) \\ = .20 + .25 + .20 \\ = .65 \end{aligned}

P (X < 3) = .10 + .15 + .20 = .45

☐

Question

Pasted image 20241003170912.png

Proof
Here we have $W \to Y^{'} = 0, T \to Y^{'} = 1, . . .$ . So then we get the Probability Distribution, PMF:

$Y^{'}$	0	1	2	3
$p (Y^{'})$	.3	.4	.2	.1
For each one. For both we have to consider all the possiblities, where we mutliply the probabilties since it's given the two magazines arriving are independent:

	$Y_{1} = 0$	$Y_{1} = 1$	$Y_{1} = 2$	$Y_{1} = 3$
$Y_{2} = 0$	.09	.12	.06	.03
$Y_{2} = 1$	.12	.16	.08	.04
$Y_{2} = 2$	.06	.08	.04	.02
$Y_{2} = 3$	.03	.04	.02	.01
Then to get the real $Y$ for the number of days for both magazines, we just do: