Probability(win game) \(\displaystyle={p}{\left({W}\right)}={0.34}\)

let \(\displaystyle{W}={w}\in,{L}={l}{o}{s}{e}\)

a) \(\displaystyle{p}{\left(\text{winning 2 wars simultaneously}\right)}={0.34}\cdot{0.34}={0.1156}\)

b) \(\displaystyle{p}{\left(\text{winning 4 games simultaneously}\right)}={0.34}\times{0.34}\times{0.34}\times{0.34}={0.34}^{{{4}}}={0.01336}\)

c)events are independent, their complements are independent as well.

A W has 0.34 probability. A L has 0.66 probability

So \(\displaystyle{p}{\left(\text{winning 4 in a row and failure in the 5th}\right)}={0.34}^{{{4}}}\times{0.66}={0.00882}\)

let \(\displaystyle{W}={w}\in,{L}={l}{o}{s}{e}\)

a) \(\displaystyle{p}{\left(\text{winning 2 wars simultaneously}\right)}={0.34}\cdot{0.34}={0.1156}\)

b) \(\displaystyle{p}{\left(\text{winning 4 games simultaneously}\right)}={0.34}\times{0.34}\times{0.34}\times{0.34}={0.34}^{{{4}}}={0.01336}\)

c)events are independent, their complements are independent as well.

A W has 0.34 probability. A L has 0.66 probability

So \(\displaystyle{p}{\left(\text{winning 4 in a row and failure in the 5th}\right)}={0.34}^{{{4}}}\times{0.66}={0.00882}\)