Question

In a lottery there were $90$ tickets numbered $1$ to $90$. Five tickets were drawn at random. The probability that two of the tickets drawn number $15$ and $89$ is-

• A. $\frac{2}{801}$
• B. $\frac{2}{623}$
• C. $\frac{1}{267}$
• D. $\frac{1}{623}$

Answer 1

Answer: (A)

$\frac{2}{801}$

$5$ ticket selected = ${}^{90}{C}_5$ (total space)

2 particular ticket selected = $1$ ways

Rest 3 tickets from $88$ tickets = ${}^{88}{C}_3$

Probability = $\frac{1.1{}^{88}{C}_3}{{}^{90}{C}_5}=\frac{\frac{88\times 87\times 86}{3\times 2}}{\frac{90\times 89\times 88\times 87\times 86}{5\times 4\times 3\times 2\times 1}}$

$P=\frac{20}{90\times 89}=\frac{2}{801}$