Question

One ticket is selected at randon from $100$ tickets numbered $00,01,02,....,99.$ Suppose A and B are the sum and product of the digits found on the ticket. Then $P\left(\right(A=7\left)/\right(B=0\left)\right)$ is given by

• A. $2/13$
• B. $2/19$
• C. $1/50$
• D. None of these

Answer 1

Answer: (B)

$2/19$

Total number of ways , $n\left(S\right)=100$

Sample space for product of digits equal to zero is = $01,02,03,04,05,06,07,08,09,10,20,30,40,50,60,70,80,90,00$

So,

$n\left(B\right)=19$

$P\left(B\right)=\frac{19}{100}$

Sample space for sum of digits equal to 7 is =$07,16,25,34,43,52,61,70$

Now, sample space for the sum and product of the digits on the tickets is =$07,70$

Now $P(A\cap B)=\frac{2}{100}$

Hence, $P\left(\frac{(A=7)}{(B=0)}\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\frac{2}{100}}{\frac{19}{100}}=\frac{2}{19}$