Question

‘n’ whole numbers are randomly chosen and multiplied, then probability that

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(a\right)& \text{The last digit is 1,3,7 or 9}& \left(p\right)& \frac{8^n-4^n}{{10}^n}\\ \left(b\right)& \text{The last digit 2,4,6,8}& \left(q\right)& \frac{5^n-4^n}{{10}^n}\\ \left(c\right)& \text{The last digit is 5}& \left(r\right)& \frac{4^n}{{10}^n}\\ \left(d\right)& \text{The last digit is zero}& \left(s\right)& \frac{{10}^n-8^n-5^n+4^n}{{10}^n}\end{array}$

• A. (A-r), (B-p), (C-q), (D-s)
• B. (A-p), (B-r), (C-q), (D-s)
• C. (A-r), (B-q), (C-s), (D-p)
• D. (A-s), (B-q), (C-p), (D-s)

Answer 1

The correct option is A

(A-r), (B-p), (C-q), (D-s)

(a) The required event will occur if last digit in all the chosen numbers is 1, 3, 7 or 9.

$\therefore$ Req. probability $={\left(\frac{4}{10}\right)}^n$

(b) Required probability = P (that the last digit is (2, 4, 6, 8) = P(That the last digit is 1, 2, 3, 4, 6, 7, 8, 9) P (that the last digit is 1, 3, 7, 9) $=\frac{8^n-4^n}{{10}^n}.$

$\left(c\right)P(1,3,5,7,9)-P(1,3,7,9)=\frac{5^n-4^n}{{10}^n}$

(d) required prob $=P(0,5)-P\left(5\right)=\frac{({10}^n-8^n)-(5^n-4^n)}{{10}^n}$

$=\frac{{10}^n-8^n-5^n+4^n}{{10}^n}.$