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A student says: 'When two coins are tossed simultaneously, there are three outcomes — two heads, two tails, or one of each — so the probability of getting one of each is $\frac{1}{3}$.' Which option correctly evaluates this argument?
ACorrect, because the three descriptions are the only possible outcomes and are equally likely.
BIncorrect, because there are actually four equally likely ordered outcomes: $(H,H)$, $(H,T)$, $(T,H)$, $(T,T)$.
CCorrect, because $(H,T)$ and $(T,H)$ are impossible.
DIncorrect, because the probability of one of each is $\frac{1}{4}$.
Answer & Solution
Correct answer: B. Incorrect, because there are actually four equally likely ordered outcomes: $(H,H)$, $(H,T)$, $(T,H)$, $(T,T)$.
The key principle is that probabilities must be assigned over equally likely elementary outcomes. When two coins are tossed, the elementary outcomes are $(H,H)$, $(H,T)$, $(T,H)$ and $(T,T)$, so there are 4 equally likely outcomes, not 3.
The event 'one of each' includes two elementary outcomes: $(H,T)$ and $(T,H)$. Therefore its probability is $\frac{2}{4}=\frac{1}{2}$, not $\frac{1}{3}$.
Option A is tempting because 'two heads', 'two tails' and 'one of each' sound like three categories, but they are not equally likely categories. Option C is false because $(H,T)$ and $(T,H)$ are valid distinct outcomes. Option D is also incorrect because 'one of each' has two favourable outcomes, not one.
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