Answer:
a) [tex]S=50[/tex]
[tex]P(X)=0.02[/tex]
b) [tex]P(W,M,C)=8*10^-^6[/tex]
c) [tex]P(W_2_3)=1.18*10^-^3[/tex]
Step-by-step explanation:
From the question we are told that
Sample space S=50
Sample size n=30
a)Generally the sample space S is
[tex]S=50[/tex]
The probability measure is given as
[tex]P(X)=\frac{1}{50}[/tex]
[tex]P(X)=0.02[/tex]
b)
Generally the probability that the class hangs Wisconsin’s flag on Monday, Michigan’s flag on Tuesday, and California’s flag on Wednesday is mathematically given as
Probability of each one being hanged is
[tex]P(X)=\frac{1}{50}[/tex]
Therefore
[tex]P(W,M,C)=\frac{1}{50} *\frac{1}{50}* \frac{1}{50}[/tex]
[tex]P(W,M,C)=\frac{1}{125000}[/tex]
[tex]P(W,M,C)=8*10^-^6[/tex]
c)Generally the probability that Wisconsin’s flag will be hung at least two of the three days is mathematically given as
Probability of two days hung +Probability of three days hung
Therefore
[tex]P(W_2_3)=^3C_2 (1/50) * (1/50) * (49/50) +^3C_3 (1/50) * (1/50) *(1/50)[/tex]
[tex]P(W_2_3)=148 / 125000[/tex]
[tex]P(W_2_3)=1.18*10^-^3[/tex]
