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Red light! Refer to Exercise 84. Calculate and interpret P(Y≥7)

Short Answer

Expert verified

Around 26.60%chances are there for the Pedro getting at least 7 red lights on the 10 randomly selected working days.

Step by step solution

01

Given Information

Number of trials, n=10

Probability of success, p=55%=0.55

Y:The number of times that the light is red.

02

Simplification

According to the binomial probability,

P(X=k)=nk·pk·(1-p)n-k

Addition rule for mutually exclusive event:

P(A∪B)=P(AorB)=P(A)+P(B)

At k=7,

The binomial probability to be evaluated as:

P(X=7)=107·(0.55)7·(1-0.55)10-7=10!7!(10-7)!·(0.55)7·(0.45)3=120·(0.55)7·(0.45)3≈0.1665

At k=8,

The binomial probability to be evaluated as:

P(X=8)=108·(0.55)8·(1-0.55)10-8=10!8!(10-8)!·(0.55)8·(0.45)2=45·(0.55)8·(0.45)2≈0.0763

At k=9,

The binomial probability to be evaluated as:

P(X=9)=109·(0.55)9·(1-0.55)10-9=10!9!(10-9)!·(0.55)9·(0.45)1=10·(0.55)9·(0.45)1≈0.0207

03

Simplificaiton

At k=10,

The binomial probability to be evaluated as:

P(X=10)=1010·(0.55)10·(1-0.45)10-10=10!10!(10-10)!·(0.55)10·(0.45)0=1·(0.55)10·(0.45)0≈0.0025

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

role="math" localid="1653975671376" P(X≥7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)=0.1665+0.0763+0.0207+0.0025=0.2660=26.60%

Thus,

Around 26.60%chances are there for the Pedro getting at least 7 red lights on the 10 randomly selected working days

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