Official solution
Coin is tossed 5 times, so n = 5
Let, p = probability of getting heads
q = probability of getting tails.
\[\therefore\] p + q = 1 ...... (1)
\[\therefore\] Probability of getting 4 heads
= 5C4 . p4 . q
And probability of getting 5 heads
= 5C5 . p5
Given, 5C4 . p4 . q = 5C5 . p5
\[\Rightarrow\] 5q = p ....... (2)
From equation (1) and (2), we get,
5q + q = 1
\[\Rightarrow\] 6q = 1
\[\Rightarrow\] q = \[{1 \over 6}\]
\[\therefore\] p = 1 - \[{1 \over 6}\] = \[{5 \over 6}\]
Now, probability of getting atmost two heads
= p (x = 0) + p (x = 1) + p (x = 2)
p (x = 0) = Getting zero head in 5 trials
= 5C0 . p0 . q5
p (x = 1) = Getting one head in 5 trials
= 5C1 . p1 . q4
p (x = 2) = Getting two heads in 5 trials
= 5C2 . p2 . q3
= 5C0 . q5 + 5C1 . pq4 + 5C2 . p2q3
\[ = {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}\]
\[ = {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}\] = \[{{46} \over {{6^4}}}\]