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30 adults booked to stay in a hotel.
19 adults booked breakfast
15 adults booked dinner
4 adults did not book breakfast or dinner
Some adults booked breakfast and dinner.
Meihui chooses at random two of the 30 adults.
Work out the probability that these two adults each booked breakfast and dinner.

$30$ adults booked to stay in a hotel. $\newline$ $19$ adults booked breakfast $\newline$ $15$ adults booked dinner $\newline$ $4$ adults did not book breakfast or dinner $\newline$ Some adults booked breakfast and dinner. $\newline$ Meihui chooses at random two of the 30 adults. $\newline$ Work out the probability that these two adults each booked breakfast and dinner.

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Full solution

Q.

30

adults booked to stay in a hotel.

\newline

19

adults booked breakfast

\newline

15

adults booked dinner

\newline

4

adults did not book breakfast or dinner

\newline

Some adults booked breakfast and dinner.

\newline

Meihui chooses at random two of the 30 adults.

\newline

Work out the probability that these two adults each booked breakfast and dinner.

Find Total Adults: First, let's find out how many adults booked both breakfast and dinner. $\newline$ Total adults = $1930$ $\newline$ Adults booked breakfast = $19$ $\newline$ Adults booked dinner = $15$ $\newline$ Adults booked neither = $4$ $\newline$ So, adults who booked either or both = $1930 - 4 = 1926$
Apply Inclusion-Exclusion Principle: Now, we use the principle of inclusion-exclusion to find the number who booked both. $\newline$ Adults who booked both = Adults booked breakfast + Adults booked dinner - Adults who booked either or both $\newline$ Adults who booked both = $19 + 15 - 1926$

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