Question

Suppose there are n type of coupons. Each new coupon collected is of type i with probability Pi, independently of any other collected coupon. Here, D=1 Pi = 1. Suppose k coupons are collected. Let A be the event that there is at least one coupon of type i among the k collected. For i #j, (1) Compute P(A|AU A;) (2) Compute P(A|Aj)

Answer 1

Answer:-

Given That:-

Suppose there are n type of coupons. Each new coupon collected is of type i with probability Pi, independently of any other collected coupon. Here, $\sum_{i=1}^{n}p_i=1$ . Suppose k coupons are collected. Let A_i be the event that there is at least one coupon of type i among the k collected. For i $\neq$ j,

(1) Compute  $P(A_i|A_i\cup A_j)$

The given conditional probability is computed using Bayes theorem here as:

$P(A_i|A_i\cup A_j)$

Here,

means at least one of the coupon type: i or j has been pulled out. the probability of this event is computed as:

A; UA

= 1 - Probability that none of the coupon types
A,
or  $A_j$ have pulled out

$= 1 - \left ( \sum _{L\neq i,j}p_L \right )^k$

Now probability that has been pulled is computed here as:

A,

= 1 - Probability that has not been pulled out in all the k trials

A,

$= 1 - \left ( \sum _{L\neq i}p_L \right )^k$

Therefore using Bayes theorem now the probability is comupted here as:

$P(A_i|A_i\cup A_j)=\frac{P(A_i)}{P(A_i\cup A_j)}$

$P(A_i|A_i\cup A_j)=\frac{1-(\Sigma _{L\neq i}p_L)^k}{1-(\Sigma _{L\neq i,j}p_{L})^k}$

(2) Compute P(A_i | A_j)

The probability here is computed using Bayes theorem as:

$P(A_i|A_j)=\frac{P(A_i \cap A_j)}{P(A_j)}$

The numerator here is the probability that both coupon types: i and j have pulled in k trials.

$P(A_i\cap A_j)=P(A_i)+P(A_j)-P(A_i \cup A_j)$

$P(A_i\cap A_j)=1-\left ( \sum _{L\neq i}p_{L} \right )^k+1-\left ( \sum _{L\neq j}p_{L} \right )^k-\left ( 1-\left ( \sum _{L\neq i,j}p_{L} \right )^k \right )$

$P(A_i\cap A_j)=1-\left ( \sum _{L\neq i}p_{L} \right )^k-\left ( \sum _{L\neq j}p_{L} \right )^k+\left ( \sum _{L\neq i,j}p_{L} \right )^k$

Now putting this, we get the final probability here as:

$P(A_i|A_j)=\frac{P(A_i \cap A_j)}{P(A_j)}=\frac{1-(\Sigma _{L\neq i}p_L)^k-(\Sigma _{L\neq j}p_L)^k+(\Sigma _{L\neq i,j}p_L)^k}{1-(\Sigma _{L\neq j}p_L)^k}$

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