Question

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

$$\begin{array}{cc}ColumnI& ColumnII\\ \left(A\right)thelastdigitis1,3,7,or9& \left(P\right)\frac{8^n-4^n}{{10}^n}\\ \left(B\right)thelastdigitis2,4,6,r8& \left(Q\right)\frac{5^n-4^n}{{10}^n}\\ \left(C\right)thelastdigitis5& \left(R\right)\frac{4^n}{{10}^n}\\ \left(D\right)thelastdigitiszero& \left(S\right)\frac{{10}^n-8^n-5^n+4^n}{{10}^n}\end{array}$$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$(A-R),(B-P),(C-Q),(D-S)$

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

Therefore required probability = $(\frac{4}{10}{)}^n$

(B) Required probability = P (that the last digit is 2,4,6,8)= P(that the last digit is 1, 2, 3, 4, 5, 6, 7, 8,9)-P(that the last digit is 1, 3, 7, 9)= $\frac{{10}^n-4^n}{{10}^n}$

(C) $P(1,3,5,7,9)-P(1,3,7,9)$

$\frac{5^n-4^n}{{10}^n}$

(D) Required probability = $P(0,5)-P\left(5\right)$

$=\frac{({10}^n-8^n)-(5^n-4^n)}{{10}^n}$

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