Problem 5.15 Easy Free 2 variations

Given At Least One Head

Three fair coins are tossed, and you learn that at least one of them landed heads. Find the probability that exactly one coin landed heads.

Work it out, then compare with the answer.

SolutionShow the worked solution

List the outcomes as ordered triples. All $2^3=8$ are equally likely. The information “at least one head” rules out only $TTT$ . Conditioning therefore leaves the remaining $7$ outcomes, and they stay equally likely. Exactly one head occurs in three of them: $HTT$ , $THT$ and $TTH$ . Hence

\mathbb{P}(\text{exactly one head}\mid\text{at least one head})=\frac{\mathbb{P}(\text{exactly one head})}{1-\mathbb{P}(TTT)}=\frac{3/8}{7/8}=\frac{3}{7}.

\boxed{\dfrac{3}{7}}

Key idea. Conditioning on “at least one” removes only the complementary outcome. The rest keep their relative weights, so the probability is a simple count.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 5.15a

At a zoo exhibit there are two species, giant otters and red pandas, and each animal present is engaged in one of five activities: eating, sleeping, rolling, swimming, or resting. You observe two animals at the exhibit, and you are told that at least one of them is a swimming red panda. Assuming every combination of species and activity (including both animals matching) is equally likely a priori, find the probability that both animals are swimming.

Work it out, then compare with the answer.

Solution

Each animal takes one of $2\times5=10$ equally likely (species, activity) states, so there are $100$ equally likely ordered pairs. Write $RS$ for a swimming red panda.

Pairs with no $RS$ on either animal number $9^2=81$ , so $100-81=19$ pairs contain at least one $RS$ .

Pairs with both animals swimming number $2\times2=4$ . Of these, all except (otter, otter) contain an $RS$ , which gives $3$ . Therefore

\mathbb{P}(\text{both swimming}\mid\text{at least one } RS)=\frac{3}{19}.

\boxed{\dfrac{3}{19}}

Variation 5.15b

A family has two children. You learn that at least one of them is a boy born on a Friday. Assuming each child’s sex and day of birth are independent and uniformly distributed, find the probability that the family has two boys.

Work it out, then compare with the answer.

Solution

Each child takes one of $2\times7=14$ equally likely (sex, weekday) states, so there are $196$ equally likely ordered families. Write $B^*$ for a boy born on a Friday.

Families with at least one $B^*$ number $14^2-13^2=27$ .

Two-boy families have $7^2$ ordered weekday choices. Of these, $6^2$ have no Friday birth, so $7^2-6^2=13$ two-boy families contain a $B^*$ . Hence

\mathbb{P}(\text{two boys}\mid\text{at least one } B^*)=\frac{13}{27}\approx0.481.

\boxed{\dfrac{13}{27}}

Without the weekday detail the answer would be $\tfrac13$ . The extra condition is far more likely to be met when there are two boys, so it pushes the answer towards $\tfrac12$ .

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