Q1: There are four electronic components in a system named A,B,C and D. Let P (A)=P(component A works)=0.6. Similarly, P (B) = 0.4, P (C) = 0.5, P (D) = 0.3. The system will work only if components A and B both work or components C and D both work. Find the probability that the system works. All components work independently.
Q2: Assume that births of boys and girls are equally likely and occur independently. Suppose that a family has seven children and it is known that there are at least 5 boys. What is the probability that there are exactly 6 boys?
My work for Q1:
$P(A \cap B)= (0.6)(0.4)=0.24$,
$P(C \cap D)= (0.5)(0.3)=0.15$
Required probability=$P((A \cap B) \cup (C \cap D))= P(A \cap B) + P(C \cap D) - P(A \cap B \cap C \cap D)$
$= 0.24 + 0.15 - (0.6)(0.4)(0.5)(0.3)$
$=0.354$ (Final answer)
My work for Q2:
$P(B)= 0.5,P(G)=0.5$
Required probability: $P(6B)| At least 5B)$ $= \frac{ P(6B \space \cap \space At least 5 B)}{P(At least 5B)}$
$= \frac{ P(6B \space \cap \space (5B \cup 6B \cup 7B))}{P(At least 5B)}$
$= \frac {P(6B)}{P(at least 5B)}$
Then $P(5B)= \binom{7}5 (0.5)^5 (0.5)^2 = \frac{21}{128}$
$P(6B)= \binom{7}6 (0.5)^6 (0.5)^1 = \frac{7}{128}$
$P(7B)= \binom{7}7 (0.5)^7 (0.5)^0 = \frac{1}{128}$
$ \frac {P(6B)}{P(at least 5B)} = \frac{P(6B)}{P(5B) +P(6B) + P(7B)} = \frac{7}{29}$ (Final answer)
Is my logic and reasoning correct for both of these questions?
Your work is nice but I would suggest an easy way to solve both the exercises (useful to save time in the solutions)
Q1:
It's easier find the complement probability that both branches do not work
$$1-0.76\times 0.85=0.354$$
Q2:
It's easier just counting the possible and favourable combinations (the probability of one single realization is always the same, thus it can be wasted in numerator and denominator)
$$\frac{\binom{7}{1}}{\binom{7}{2}+\binom{7}{1}+\binom{7}{0}}=\frac{7}{21+7+1}=\frac{7}{29}$$