Q.A bag contains 5 red and 3 black balls. Two balls are drawn one after another, WITHOUT replacement. Find the probability that both balls drawn are red.

Condition on first $=$ girl: sample space $\{GG,GB\}$; $P(\text{both girls})=\tfrac12$.

Let $G$ = girl, $B$ = boy, with the first letter denoting the elder child. The full sample space is $\{GG,\ GB,\ BG,\ BB\}$.

We are given that the first child is a girl, so the reduced (conditional) sample space is

$\{GG,\ GB\},$ …

$P(A \cap B) = \tfrac35 - \tfrac12 = \tfrac{1}{10}$; hence $P(A/B) = \dfrac{1/10}{3/5} = \tfrac16$ and $P(A' \cup B') = 1 - \tfrac{1}{10} = \tfrac{9}{10}$.

GSEB Class-12 Statistics, Probability (conditional probability):

Step 1 — find $P(A \cap B)$. The event $B$ splits into $A \cap B$ and $A' \cap B$:

$P(B) = P(A \cap B) + P(A' \cap B)$

$P(A \cap B) = P(B) - P(A' \cap B) = \frac{3}{5} - \frac{1}{2} = \frac{6 - 5}{10} = \frac{1}{10}$

Step 2 — conditional probability $P(A/B)$: …

$P(A)=1-\frac{7}{25}=\frac{18}{25}$; $P(A\cap B)=\frac{18}{25}\cdot\frac{5}{12}=\frac{3}{10}$; $P(B)=\frac{P(A\cap B)}{P(A/B)}=\frac{3/10}{1/2}=\frac{3}{5}$.

Given $P(A')=\dfrac{7}{25}$, $P(B/A)=\dfrac{5}{12}$, $P(A/B)=\dfrac12$.

Step 1 — $P(A)$:

$P(A)=1-P(A')=1-\frac{7}{25}=\frac{18}{25}.$

Step 2 — $P(A\cap B)$ by the multiplication theorem $P(A\cap B)=P(A)\cdot P(B/A)$:

$P(A\cap B)=\frac{18}{25}\times\frac{5}{12}=\frac{90}{300}=\frac{3}{10}=0.3.$

…

Room dark only if both bulbs defective: $P = \frac{3}{10}\cdot\frac{2}{9} = \frac{1}{15}$; $P(\text{lit}) = 1 - \frac{1}{15} = \frac{14}{15}$. (OR: three Venn diagrams described.)

Main part. The box has 10 bulbs, of which 3 are defective and 7 are good. Two bulbs are drawn at random (without replacement) and fixed in two holders. The room is lighted if at least one of the two bulbs is good; it stays dark only if both are defective.

$P(\text{both defective}) = \frac{3}{10}\times\frac{2}{9} = \frac{6}{90} = \frac{1}{15}$

By the complement rule:

$P(\text{room lighted}) = 1 - P(\text{both defective}) = 1 - \frac{1}{15} = \frac{14}{15} \approx 0.9333$

OR — Venn diagrams (drawn inside a rectangle representing the sample space $S$):

1. Mutually exclusive events $A$ and $B$: two circles that do not overlap (no common region), since $A \cap B = \varnothing$. …