Probability of HH in two tosses.

**The reasoning** When you toss a coin twice, you're looking at **independent events** — each toss doesn't affect the other. List all possible outcomes: - HH (heads, then heads) - HT (heads, then tails) - TH (tails, then heads) - TT (tails, then tails) That's **4 equally likely outcomes**. Only **1 of them is HH**. So: P(HH) = Number of favorable outcomes ÷ Total outcomes = 1/4 Or use the **multiplication rule**: P(H on first toss) × P(H on second toss) = 1/2 × 1/2 = **1/4** **Why the wrong options tempt you** **B) 1/2** — You might think "either I get HH or I don't," but that ignores that HT, TH, and TT are three *different* ways to "not get HH." **C) 3/4** — Confusing this with "probability of at least one H" (which is indeed 3/4). **D) 1** — Thinking it's certain, but nothing's guaranteed in probability! **Quick takeaway** With two coin tosses, there are always **four** equally likely outcomes — write them out, count your target, and divide.