Probability of passing the 3rd exam

Question

Let $A_i$ be the event that the student passes the ith exam. The probability of passing the 1st exam is 0.80. If he passes the 1st exam then the chance of passing the second is 0.81 but if fails the first then the chance he passes the second drops to 0.40. If he passes both then his chance of passing the third exam is 0.82. If he only passes one of the first two then the chance of passing the third is 0.60. If he fails both then the chance of passing the 3rd is 0.10.

1. Probability of passing the 1st and 2nd exam: $P(A_1A_2)=P(A_2|A_1)P(A_1)=(0.81)(0.80)=0.648$
2. Probability of failing the 1st and 2nd exam: $P(A_1^CA_2^C)=P(A_2^C|A_1^C)P(A_1^C)=(1-0.80)(1-0.40)=0.12$
3. Probability of failing the 1st but passing the 2nd: $P(A_1^CA_2)=P(A_2|A_1^C)P(A_1^C)=(0.40)(0.20)=0.08$
4. Probability of passing the 3rd exam: $P(A_3)=P(A_3|A_1A_2)P(A_1A_2)+P(A_3|A_1^C \cup A_2^C)P(A_1^C \cup A_2^C)$

I'm struggling with #4, I'm not sure if I have the right idea in my attempt since the $A_i$'s are not mutually exclusive and if my attempt is correct, I still don't know how to go about finding $P(A_3|A_1^C \cup A_2^C)$. Please let me know how I could do #4 and if anything needs to be corrected.

Answer 1

For #4,

$P(A_3) = P(A_3|A_1A_2)P(A_1A_2)+P(A_3|A_{1}^{C}A_2)P(A_{1}^{C}A_2)+P(A_3|A_1A_{2}^{C})P(A_1A_{2}^{C})+P(A_3|A_{1}^{C}A_{2}^{C})P(A_{1}^{C}A_{2}^{C})$

$= 0.82P(A_1A_2)+0.60P(A_{1}^{C}A_2)+0.60P(A_1A_{2}^{C})+0.10P(A_{1}^{C}A_{2}^{C})$

$=(0.82\times 0.648)+(0.60 \times 0.08)+(0.60 \times 0.152)+(0.10 \times 0.12)$

$= 0.683$

Answer 2

By using TREE DIAGRAM

$$\text{P(passing 3rd exam)}=0.8\times 0.81\times 0.82 + 0.8\times 0.19\times 0.6 + 0.2\times 0.4\times 0.6 + 0.2\times 0.6\times 0.1 = 0.68256$$

Answer 3

We are given $$\Pr[A_1] = 0.8, \quad \Pr[\bar A_1] = 1 - 0.8 = 0.2; \\ \Pr[A_2 \mid A_1] = 0.81, \quad \Pr[A_2 \mid \bar A_1] = 0.40; \\ \Pr[A_3 \mid (A_1 \cap A_2)] = 0.82, \quad \Pr[A_3 \mid (A_1 \cap \bar A_2) \cup (\bar A_1 \cap A_2)] = 0.6, \quad \Pr[A_3 \mid (\bar A_1 \cap \bar A_2)] = 0.10.$$

For the sake of simplicity, let us denote $X \in \{0,1,2\}$ the number of tests that are passed among the first two tests. So we may write more compactly $$\Pr[A_3 \mid X = 2] = 0.82, \quad \Pr[A_3 \mid X = 1] = 0.6, \quad \Pr[A_3 \mid X = 0] = 0.1.$$

Then the law of total probability gives $$\Pr[A_3] = \sum_{x=0}^2 \Pr[A_3 \mid X = x]\Pr[X = x].$$ All that remains is to determine $\Pr[X = x]$ for each $x \in \{0, 1, 2\}$. To this end, we note $$\Pr[X = 2] = \Pr[A_1 \cap A_2] = \Pr[A_2 \mid A_1]\Pr[A_1] = (0.81)(0.8).$$ We also have $$\Pr[X = 1] = \Pr[A_1 \cap \bar A_2] + \Pr[\bar A_1 \cap A_2],$$ since each event (pass one test but not the other) is mutually exclusive of the other. Then $$\Pr[X = 1] = \Pr[\bar A_2 \mid A_1]\Pr[A_1] + \Pr[A_2 \mid \bar A_1]\Pr[\bar A_1] = (1 - 0.81)(0.8) + (0.4)(0.2).$$

Finally, $$\Pr[X = 0] = \Pr[\bar A_1 \cap \bar A_2] = \Pr[\bar A_2 \mid \bar A_1]\Pr[\bar A_1] = (1 - 0.4)(0.2).$$

So the desired unconditional probability is just

$$\Pr[A_3] = (0.82)(0.81)(0.8) + (0.6)((0.19)(0.8) + (0.4)(0.2)) + (0.1)(0.6)(0.2).$$