Let a biased coin be tossed 5 times. If the probability of getting...

$P\left(4heads\right) = P\left(5heads\right)$

${}^{5}{C}_{4}{p}^{4}q = {}^{5}{C}_{5}{p}^{5} ⇒ p = 5q . . . \left(i\right)$

Also, $p + q = 1 ⇒ q = \dfrac{1}{6}, p = \dfrac{5}{6}$ (from \left(i\right)$)</p><p>So,$P$(at most$2$heads)$= P\left(x = 0\right) + P\left(x = 1\right) + P\left(x = 2\right)$</p><p>$= {q}^{5}{+}^{5}{C}_{1}p{q}^{4}{+}^{5}{C}_{2}{p}^{2}{q}^{3}$</p><p>$= {\left(\dfrac{1}{6}\right)}^{5} + 5$\left(\dfrac{5}{6}\right){\left(\dfrac{1}{6}\right)}^{4} + 10{\left(\dfrac{5}{6}\right)}^{2}{\left(\dfrac{1}{6}\right)}^{3}$ $= \dfrac{46}{{6}^{4}}$