Bayes with sampling without replacement

2 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.

1. Find the probability that they are both toffees.
2. Given that they are both toffees, find the probability that they both came from box \(A\).

1. A box contains only red counters and black counters.

There are \(n\) red counters and \(n + 1\) black counters.
Two counters are selected at random, one at a time without replacement, from the box.

1. Complete the tree diagram for this information. Give your probabilities in terms of \(n\) where necessary. \includegraphics[max width=\textwidth, alt={}, center]{fe416f2e-bc81-444b-a0ca-f0eae9a8b149-24_940_1180_591_413}
2. Show that the probability that the two counters selected are different colours is $$\frac { n + 1 } { 2 n + 1 }$$ The probability that the two counters selected are different colours is \(\frac { 25 } { 49 }\)
3. Find the total number of counters in the box before any counters were selected. Given that the two counters selected are different colours,
4. find the probability that the 1st counter is black. You must show your working.

The values of the two variables \(A\) and \(B\) given in the table in Question 5 are written on cards and placed in two separate packs, which are labelled \(A\) and \(B\). One card is selected from Pack \(A\). Let \(A_i\) represent the event that the first digit on this card is \(i\).

1. Write down the value of P\((A_2)\). [1 mark] The card taken from Pack \(A\) is now transferred into Pack \(B\), and one card is picked at random from Pack \(B\). Let \(B_i\) represent the event that the first digit on this card is \(i\).
2. Show that P\((A_1 \cap B_1) = \frac{1}{24}\). [3 marks]
3. Show that P\((A_6 | B_2) = \frac{4}{41}\). [5 marks]
4. Find the value of P\((A_1 \cup B_4)\). [5 marks]