P(X< 3 n=15, successes=.4)
Think of a string of letters 15 places long.
“FFFFFFFFFFFFFFF” that would be a string of 15 failures and can happen in only with probability $\displaystyle (0.6)^{15}$. That is because P(success) $\displaystyle =0.4$ while P(failure) $\displaystyle ={1-(0.4)}=0.6$.
“SFFFFFFFFFFFFFF” that would be a string of 14 failures and one success. It can happen in 15 ways (S can be in 15 places). $\displaystyle P(X = 1) = 15\left( {.04} \right)\left( {0.6} \right)^{14} $.
“SSFFFFFFFFFFFFF” that would be a string of 13 failures and two successes. It can happen in combination of 15 taken 2 at a time ways. $\displaystyle P(X = 2) = {{15} \choose 2}\left( {.04} \right)^2 \left( {0.6} \right)^{13} $
$\displaystyle P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$