Question

27. Boxes of sweets contain toffees and chocolates. Box A contains 6 toffees and 4 chocolates, box B contains 5 toffees and 3 chocolates, and box C contains 3 toffees and 7 chocolates. One of the boxes is chosen at random and two sweets are taken out, one after the other, and eaten. (i) Find the probability that they are both toffees. (ii) Given that they are both toffees, find the probability that they both came from box A

Answer 1

The probability that both sweets are toffees is 7/20. Given that they are both toffees, the probability that they both came from Box A is 8/7.

Step-by-step explanation:

To find the probability that both sweets are toffees, we need to find the probability of selecting a toffee from each box and then add them up. Box A contains 6 toffees out of 10 sweets, so the probability of selecting a toffee from Box A is 6/10. Similarly, Box B contains 5 toffees out of 8 sweets, so the probability of selecting a toffee from Box B is 5/8. Box C contains 3 toffees out of 10 sweets, so the probability of selecting a toffee from Box C is 3/10. Now, since we are choosing a box at random, the probability of choosing any particular box is 1/3. To find the overall probability, we multiply the probability of choosing a particular box by the probability of selecting a toffee from that box. Therefore, the probability of selecting toffees from both boxes is (1/3) * (6/10) + (1/3) * (5/8) + (1/3) * (3/10) = 1/5 + 5/24 + 1/10 = 7/20.

To find the probability that the toffees came both from Box A, given that they are both toffees, we need to find the conditional probability. The conditional probability is the probability of the intersection of the two events divided by the probability of the first event. In this case, the intersection of the two events is the event that both toffees came from Box A, and the first event is the event that both sweets are toffees. Therefore, the conditional probability is (1/3) * (6/10) / (7/20) = (2/5) / (7/20) = 8/7.

Answer 2

Where the above scenario is given:

Probability that they are both Toffess is 53/210

Probability that they are both from Box A is 70/159

Let E1 : box A is selected

E2: Box B is selected
E3: Box C is selected

A: Two toffes are selected:

Since these are three boxes and one box is selected at random:

Hence, P (E1) = 1/3, P (E2) = 1/3 and P E3) = 1/3

Now Box A contains 6 Toffees and 4 chocolates.

Hence, P (A | E1) = 6/10 x 5/9 = 30/90 = 1/3

Box B contains 5 Toffees and 3 chocolates

Thus, P (A|E2_ = 5/8 x 4/7 = 20/56 = 5/ 14

Box C contains 3 toffess and 7 chocolates

Thus, P (A|E3) = 3/10 x 2/9 = 1/15


(!) Now by law of the total probability,

P(A) = P(E1) P(A|E1) + P (E2) P(A|E2) + { (E3) P (A|E3)

= 1/3 x 1/3 + 1/3 x 5/14 + 1/3 x 1/15
= 1/3 [ 1/3 + 5/4 + 1/15]
= 1/3 x (70 + 75 + 14)/210
= 1/3 x 159/210
= 53/210

hence P(A) = 53/210

Probability that they are both toffees = 53/210


II) Given that both are toffees, probabilty that they are both from Box A is :

P (E1|A)
= P (E1) P (A|E1)/ P (A) ....by Bayes Theorem
= [1/3 x 1/3] / [53/210]

= 1/9 x 210/53
= 70/159