Question

‘n’ whole numbers are randomly chosen and multiplied, then probability that
$\begin{array}{cc}Column-I& Column-II\\ \left(A\right)thelastdigitis1,3,7or9& \left(P\right)\frac{8^n-4^n}{{10}^n}\\ \left(B\right)thelastdigit2,4,6,8& \left(Q\right)\frac{5^n-4^n}{{10}^n}\\ \left(C\right)thelastdigitis5& \left(R\right)\frac{4^n}{{10}^n}\\ \left(D\right)thelastdigitiszero& \left(S\right)\frac{{10}^n-8^n-5^n+4^n}{{10}^n}\end{array}$

• A. $A\to R,B\to p,C\to q,D\to s$
• B. $A\to q,B\to r,C\to s,D\to P$
• C. $A\to q,B\to s,C\to r,D\to P$
• D. $A\to p,B\to s,C\to r,D\to q$

Answer 1

The correct option is A $A\to R,B\to p,C\to q,D\to s$

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