Box I contains 10 cards numbered 1 to 10 and Box II contains 5 cards numbered 1 to 5. One of the boxes is chosen at random and a card is drawn at random from it. The probability that the number on the card is even is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{9}{20}\)
Let \(E_1\) and \(E_2\) be the events that Box I and Box II are chosen, and let A be the event that the number drawn is even.
\( \displaystyle P(E_1) = P(E_2) = \frac{1}{2} \)
Box I has the even numbers 2, 4, 6, 8, 10 and Box II has 2, 4:
\( \displaystyle P(A|E_1) = \frac{5}{10} = \frac{1}{2} \)
\( \displaystyle P(A|E_2) = \frac{2}{5} \)
By the theorem of total probability:
\( P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \)
\( \displaystyle P(A) = \frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{2}{5} \)
\( \displaystyle P(A) = \frac{1}{4} + \frac{1}{5} = \frac{9}{20} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{7}{15}\) — Incorrect. This puts all 15 cards in one pile (7 even cards out of 15). The box is chosen first, so each card of the smaller Box II is more likely to be drawn than a card of Box I.
Option (2) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the chance of an even number from Box I alone.
Option (4) \(\displaystyle\frac{9}{10}\) — Incorrect. This adds \(\frac{1}{2}\) and \(\frac{2}{5}\) without multiplying each by the chance \(\frac{1}{2}\) of choosing that box.