Respuesta :
This kind of problems are solved using Bernoulli's distribution. Everytime you have a win/lose scenario, and you know the probability [tex] p[/tex] of winning, and you want [tex]k[/tex] successes over [tex]n[/tex] trials, you have the following probability:
[tex]\displaystyle P(k\text{ successes over }n\text{ trials}) = \binom{n}{k}p^k(1-p)^{n-k} [/tex]
You want the probability of having at least three successes, i.e. you are interested in the cases k=3 and k=4. The corresponding probabilities are
[tex]\displaystyle P(3 \text{ successes}) = \binom{4}{3}0.6^3 \cdot 0.4 = 4\cdot 0.216 \cdot 0.4 = 0.3456[/tex]
[tex]\displaystyle P(4 \text{ successes}) = \binom{4}{4}0.6^4 = 0.1296 [/tex]
So, the total probability is [tex] 0.3456+0.1296 = 0.4752[/tex]
