Lea is playing with two boxes: a White box and a Black box. The White box contai
ID: 3269547 • Letter: L
Question
Lea is playing with two boxes: a White box and a Black box. The White box contains two fair dices (each number
1, 2,..., 6 appears with probability 1/6 in each dice). The Black box contains two dices that land with probability
1/6 in each of their sides, but, in contrast to the dices in the other box, both these dices have 3 sides with a 1 and
3 sides with a 6.
Lea rolls two dice from one of the boxes, but we don’t know which box. We only know that she is
playing with the Withe and Black boxes with probability 2/3 and 1/3, respectively.
Let A1 be the event that the
first dice she rolls is a 6:{d1 = 6}, and let A2 be the event that the second dice she rolls is a 6: {d2 = 6}.
(a) (Level A) What is the probability of A2? (Hint: you may use TTP)
(b) (Level A) What is the probability of A2, given that we know that Lea is playing with the dice in the Black
box?
(c) (Level B) What is the probability of A2, given that we know that Lea is playing with the dice in the Black box
and we also know that the first dice is a 6?
Explanation / Answer
Here we are given that she is playing with the White and Black boxes with probability 2/3 and 1/3, respectively.
Therefore P(W) = 2/3 and P(B) = 1/3
A1: First roll of the dice is a 6
A2: Second roll on the dice is a 6
a) Probability of A2 would be:
= Probability that white box is selected * Probability that second roll is a 6 given that white box is selected +
Probability that black box is selected * Probability that second roll is a 6 given that black box is selected
= (2/3)*(1/6) + (1/3)(1/2)
because in white box, there are fair dice therefore probability of getting a 6 is 1/6 but for black box the probability of getting a 6 is 1/2 as there is an equal chance of getting a 1 and a 6 in that case.
Therefore we get:
= 1/9 + 1/6
= 5/18
= 0.2778
Therefore the required probability here is 0.2778
b) Probability of A2 given that Lea is playing with the dice in the black box.
= Probability that black box is selected * Probability that second roll is a 6 given that black box is selected / Probability that black box is selected
= (1/3)(1/2) / (1/3)
= (1/2)
Therefore the required probability here is 0.5
c) The number in the first dice wont have an affect on the probability of second roll. ( because the events are independent )
Therefore the required probability here is 0.5
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