Three fair coins are tossed. $E$ : the first coin shows a head. $F$ : exactly two coins show heads. Are $E$ and $F$ independent?

Yes: the three coins are tossed independently of one anotherCorrect answerYour answerYes:

P(E \cap F) = P(E)\,P(F) = \frac{1}{2} \times \frac{3}{8} = \frac{3}{16}

Correct answerYour answerNo: they are mutually exclusive, so they cannot happen togetherCorrect answerYour answerNo:

P(E \cap F) = \frac{1}{4}

but

P(E)\,P(F) = \frac{3}{16}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$P(E) = \frac{1}{2}$ and $P(F) = \frac{3}{8}$ ( $HHT, HTH, THH$ ).
$E \cap F = \{HHT, HTH\}$ : $\frac{2}{8} = \frac{1}{4} \ne \frac{1}{2} \times \frac{3}{8}$ . Not independent.

Worth noting

Independent tosses do not make every pair of events independent: knowing the first is a head raises the chance of exactly two heads from $\frac{3}{8}$ to $\frac{1}{2}$ .

Concept: Testing for independence
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsIndependent events

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Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Three fair coins are tossed. EEE: the first coin shows a head. FFF: exactly two coins show heads. Are EEE and FFF independent?

Explanation

Three fair coins are tossed. $E$ : the first coin shows a head. $F$ : exactly two coins show heads. Are $E$ and $F$ independent?