i've been practicing probability since it's not my strength, but i am doing that without a tutor or an official course, just books and videos. I was reading a problem, and i was capable of draw the Probability tree (The event and its probability given the parent branch), can you help me to find out the following probabilities, thank you very much!!
a. $P(P)$
b. $P(G)$
c. $P(B | \neg P) = P(B | NP)$
d. $P(G | P)$
e. $P(C | G)$
I guess i would need to apply the total probability theorem and Bayes Rule, but i want to supervise my steps. Thank you again.
a) $P(P) = P(P\;\cap\;A)+P(P\;\cap\;B)+P(P\;\cap\;C)$
$\quad \quad \quad = (0.25)(0.4)+(0.45)(0.2)+(0.3)(0.3) = 0.28$
b) $P(G)=P(A\;\cap\;P\; \cap G)+P(A\;\cap\;NP\; \cap G)$
$\quad \quad \quad \quad + P(B\;\cap\;P\; \cap\;G) + P(B\;\cap\;NP\;\cap\;G) $
$\quad \quad \quad \quad + P(C\;\cap\;P\; \cap\;G)+P(C\;\cap\;NP\;\cap\;G)$
$\quad \quad \quad = (0.25)(0.4)(0.85) + (0.25)(0.6)(0.2) $
$\quad\quad\quad \;\; + (0.45)(0.2)(0.65) + (0.45)(0.8)(0.1) $
$\quad\quad\quad\;\; + (0.3)(0.3)(0.1) + (0.3)(0.7)(0.3)$
$\quad\quad\quad = 0.2815$
c) $P(NP) = P(NP\;\cap\;A) + P(NP\;\cap\;B) + P(NP\;\cap\;C)$
$\quad \quad\quad\quad \quad = (0.6)(0.25) + (0.8)(0.45) + (0.7)(0.3)$
$\quad\quad\quad \quad\quad= 0.72$
$\quad P(B|NP)=\frac{P(B \;\cap\; NP)}{P(NP)} = \frac{(0.45)(0.8)}{0.72}$
d) $P(G|P)=\frac{P(G\;\cap\;P)}{P(P)}= \frac{P(G\;\cap\;P\;\cap\;A)+P(G\;\cap\;P\;\cap\;B)+P(G\;\cap\;P\;\cap\;C)}{P(P)}$
$\quad\quad\quad\quad\quad =\frac{(0.25)(0.4)(0.85) + (0.45)(0.2)(0.65) + (0.3)(0.3)(0.9)}{0.28}$
e) $P(C|G)=\frac{P(C\;\cap\;G)}{P(G)}$
$\quad\quad\quad\quad\;\; =\frac{P(C\;\cap\;G\;\cap\;P) + P(C\;\cap\;G\;\cap\;NP)}{P(G)}$
$\quad\quad\quad\quad\;\; = \frac{(0.3)(0.3)(0.9)+(0.3)(0.7)(0.3)}{0.2815}$