MHT CET · Maths · Permutation Combination

If the four positive integers are selected randomly from the set of positive integers, then the probability that the number \(1,3,7\) and 9 are in the unit place in the product of 4 -digit, so selected is

A \(\frac{7}{625}\)
B \(\frac{2}{5}\)
C \(\frac{5}{625}\)
D \(\frac{16}{625}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{16}{625}\)

Step-by-step Solution

Detailed explanation

The number of digits on unit place of any number \(=10\)
\(\therefore\) \(n(S)=10\)
The necessary condition for becoming the digits \(1,3,5\) or 7 at the unit place of product of four numbers that the digits \(1,3,5\) or 7 at unit place of every number.
\(\therefore n(A)=4\)
\(\therefore P(A)=\frac{4}{10}=\frac{2}{5}\)
So, required \(y=\left(\frac{2}{5}\right)^{4}=\frac{16}{625}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

\(\mathrm{X}=x_i\)	0	1	2	3	4
\(\mathrm{P}\left(\mathrm{X}=x_i\right)\)	\(0.4\)	\(0.3\)	\(0.1\)	\(0.1\)	\(0.1\)

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