In $x = 33^{n},n$ is a positive integral value, then what is the probability that $x$ will have 3 at its units place?

Question

Dr. Abubacker · Accepted Answer

(b)
Given that, $x = 33^{n}$
Where, $n$ is a positive integral value.
Here, only four digits may be at the unit place $ie,\ 1,\ 3,\ 7,\ 9.$
$	herefore n(S) = 4$
Let $E$ be the event of getting 3 at its units place.
$n(E) = 1$
$	herefore P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$

Q. In $x = 33^{n},n$ is a positive integral value, then what is the probability that $x$ will have 3 at its units place?

Explanation

Reviewed by Dr. Abubacker

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