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Two coins are flipped, if at least one them lands heads up, what is the probability that both do?

I've tried solving this problem countless different ways but I can never get it right. I know that "at least one" means "one or all" so I use P(AorB) then the result I divide it over the total number of ways the event can occur.

Jose
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3 Answers3

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Intuitive Answer:

There are $3$ options:

  • $HH$
  • $HT$
  • $TH$

Only $1$ of these options is $HH$, hence the probability is $\frac13$.

Formal Answer:

Let $A$ denote the event in which exactly two of the coins landing on $H$.

Let $B$ denote the event in which at least one of the coins landing on $H$.

The probability that event $A$ has happened given that event $B$ has happened:

$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1/4}{3/4}=\frac13$$

barak manos
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Let $A=$Both are head, $B=$At least one is head.

Then you are asked to find $\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}$.

Sample space=$\{HH,HT,TH,TT\}$. Then $B=\{HT,TH,HH\}$. So $P(B)=3/4$.

$A\cap B=\{HH\}$, so $P(A\cap B)=1/4$. So,$\displaystyle P(A/B)=1/3$.

Extremal
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I've tried solving this problem countless different ways but I can never get it right. I know that "at least one" means "one or all" so I use P(AorB) then the result I divide it over the total number of ways the event can occur.

Not quite so.   You are after $\mathsf P(A\cap B\mid A\cup B)$, where $A,B$ are the events of obtaining heads on each coin, respectively.   Since $(A\cap B)\subset (A\cup B)$ just use the definition of condition.

$$\mathsf P(A\cap B\mid A\cup B) = \dfrac{\mathsf P(A\cap B)}{\mathsf P(A\cup B)}$$

Remark: $\cap$ is the set intersection ("and") operator, and $\cup$ is the set union ("or") operator.

$\mathsf P(A\cap B)$ is the probability that both coins are heads.

$\mathsf P(A\cup B)$ is the probability that at least one is.

$\mathsf P(A\cap B\mid A\cup B)$ is the probability that both coins are heads when given that at least one is.

Graham Kemp
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