A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most two tails. then P(A U B) is:
Step-by-step Solution:
A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most two tails. Find \(P(A\cup B)\).
Step 1 — Sample space.
For three fair coin tosses the sample space has \(2^{3}=8\) equally likely outcomes:
\[
\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}.
\]
Step 2 — Describe events A and B.
\[
A=\{\text{exactly 2 heads}\}=\{HHT, HTH, THH\},\quad |A|=3.
\]
\[
B=\{\text{at most 2 tails}\}=\{\text{outcomes with } \#\text{tails}\le 2\}.
\]
The only outcome with more than 2 tails is \(TTT\), so
\(|B|=8-1=7\) and
\[
B=\{HHH, HHT, HTH, THH, HTT, THT, TTH\}.
\]
Step 3 — Relation between A and B.
Every outcome in A has exactly one tail, so \(A\subseteq B\). Therefore
\[
A\cup B = B.
\]
Final answer: \[ P(A\cup B)=P(B)=\frac{|B|}{8}=\frac{7}{8}. \]