Question

A box contains $10$ tickets numbered from $1$ to $10$. Two tickets are drawn one by one without replacement. The probability that the difference between the first drawn ticket number and the second is not less than $4$ is

• A. $\frac{7}{30}$
• B. $\frac{4}{30}$
• C. $\frac{11}{30}$
• D. $\frac{10}{30}$

Answer 1

The correct option is A $\frac{7}{30}$

10 tickets numbered from 1 to 10.

total no. of ways of selecting two cards $=2!{10}_{C_2}=45\times 2$

without replacement $=90$

total no. of ways of selecting two cards.

such that difference between first ticket.

numbers and second is not less than $4=$

(a, b) represents (second ticket number, first ticket number)

Ways $=(1,5)(2,5)(3,7)(4,8)(5,9)(6,10)(1,6)(2,7)(3,8)(4,9)(5,10)(1,7)$

$(2,8)(3,9)(4,10)(1,8)(2,9)(3,10)(1,9)(2,10)(1,10)$

total ways $=21$

probability $=\frac{21}{90}=\frac{7}{30}$

A is correct