Respuesta :
Answer:
Probability that it wins at least 3 of its final 5 games = .02387
Step-by-step explanation:
Given -
The probability of win the weekend game = 0.5
The probability of loose the weekend game = 0.5
If he wins the game this weekend then it will play its final 5 games in the upper bracket of its league
In this case, probability of success is (p) = 0.3
probability of failure is (q) = 1 - p = 0.7
Let X be number of game won out of last five games
probability that it wins at least 3 of its final 5 games
( 1 )
[tex]P(X\geq3)[/tex] = [tex]P(X\geq3/first\; game\; won)[/tex] ( probability of first game won )
= [tex]0.5\times[/tex]P( X =3 ) + [tex]0.5\times[/tex]P( X =4) + [tex]0.5\times P(X = 5)[/tex]
= [tex]0.5\times\binom{5}{3}(0.3)^{3}(0.7)^{2} + 0.5\times\binom{5}{4}(0.3)^{4}(0.7)^{1}[/tex] + [tex]0.5\times\binom{5}{5}(0.3)^{5}(0.7)^{0}[/tex]
= [tex]0.5\times\frac{5!}{(3!)(2!)}\times(0.3)^{3}\times(0.7)^{2} + 0.5\times\frac{5!}{(4!)(1!)}\times(0.3)^{4}\times(0.7)^{1}[/tex] + [tex]0.5\times\frac{5!}{(5!)(0!)}\times(0.3)^{5}\times(0.7)^{0}[/tex]= = .065 + .014 + .001215 = .080
If he loose the game this weekend then it will play its final 5 games in the lower bracket of its league
In this case, probability of success is (s) = 0.4
probability of failure is (t) = 1 - s = 0.6
( 2 )
[tex]P(X\geq3/first\; game\; lost)[/tex] ( probability of first game lost )
= [tex]0.5\times P(X = 3) + 0.5\times P(X = 4)[/tex] + [tex]0.5\times P(X=5)[/tex]
= [tex]\binom{5}{3}(0.4)^{3}(0.6)^{2} + 0.5\times\binom{5}{4}(0.4)^{4}(0.6)^{1}[/tex]+ [tex]0.5\times\binom{5}{5}(0.4)^{5}(0.6)^{0}[/tex]
= [tex]0.5\times\frac{5!}{(3!)(2!)}\times(0.4)^{3}\times(0.6)^{2} + 0.5\times\frac{5!}{(4!)(1!)}\times(0.4)^{4}\times(0.6)^{1}[/tex] + [tex]0.5\times\frac{5!}{(5!)(0!)}\times(0.4)^{5}\times(0.6)^{0}[/tex] = = .1152 + .0384 + .00512 = .1587
Required probability = ( 1 ) + ( 2 ) = .02387
