ProbabilityHardBloom L3

Question

Each packet of blades sold contains a coupon which is equally likely to bear the letters A, B or C. If 'm' packets are purchased, what is the probability that the coupons can not be to used to spell BAC .

Options

A.$\left( \frac{2}{3} \right)^{m}$
B.$\frac{{3.2}^{m} - 2}{3^{m}}$
C.$\frac{2^{m}}{3^{m - 1}}$
D.$\frac{2^{m} - 1}{3^{m - 1}}$

Solution

Sol. [Key Concept: Coupons can't spell BAC: inclusion–exclusion gives $\frac{3 \cdot 2^{m} - 3}{3^{m}}$.]

$$P = \frac{\ ^{3}C_{1}2^{m} - \ ^{3}C_{2}(1)^{m}}{3^{m}} = \frac{2^{m} - 1}{3^{m - 1}} $$$A \rightarrow$ number of ways s.t. there is no A

$B \rightarrow$ number of ways s.t. there is no B

$C \rightarrow$ number of ways s.t. there is no C

$$n(A \cup B \cup C) = S_{1} - S_{2} + S_{3} = \ ^{3}C_{1}2^{m} - \ ^{3}C_{2}1^{m} + 0$$

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