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Alice plays the following game on a math show. There are $7$ boxes and identical prizes are hidden inside $3$ of the boxes. Alice is asked to choose a box where a prize might be. She chooses a box uniformly at random. From the unchosen boxes which do not have a prize, the host opens an arbitrary box and shows Alice that there is no prize in it. The host then allows Alice to change her choice if she so wishes. Alice chooses a box uniformly at random from the other $5$ boxes (other than the one she chose first and the one opened by the host). Her probability of winning the prize is

1. $3 / 7$
2. $1 / 2$
3. $17 / 30$
4. $18 / 35$
5. $9 / 19$

Either the box picked by Alice at the first attempt may have had the prize or may not have had the prize, there are two event here.

Event $E1$: Alice picked the box with the prize on the first attempt which has a probability of $3/7$.

Now on the second attempt, she has a $2/5$ probability of winning the prize.

Event $E2$: Alice didn't pick the box with the prize on the first attempt which has a probability of $4/7$.

Now on the second attempt, she has a $3/5$ probability of winning the prize.

The probability of winning prize $P(W)$ is given as

$P(W) = P(E1)P(W/E1) + P(E2)P(W/E2)$

$=\left ( 3/7\right )*\left(2/5\right) + \left ( 4/7 \right )*\left ( 3/5 \right )$ = $18/35$

Hence, option D is the correct answer.