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2 years ago

P(S)= 1/7 P(T)= 0.6 If S and T are mutually exclusive events, find P(S or T).

Mathematics

1 answer:

Harlamova29_29 [7]2 years ago

3 0

Answer:

26/35

Step-by-step explanation:

If the events are mutually exclusive, all you have to do for P(S or T) is do P(S)+P(T).

So we are doing 1/7 + 0.6.

I prefer the answer as a fraction so I'm going to rewrite 0.6 as 6/10=3/5.

So we are going to add 1/7 and 3/5.

We need a common denominator which is 35.

Multiply first fraction by 5/5 and second fraction by 7/7.

We have 5/35+21/35.

This gives us 26/35.

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2 years ago

Sally can paint a room in 7 hours while it takes Steve 6 hours to paint the same room. How long would it take them to paint the

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3 years ago

The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/h

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Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

$P(X=4)=\frac{\lambda^{X}*e^{-\lambda} }{X!} =\frac{5^{4}*e^{-5} }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754$

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))$

Using

$P(X)=\frac{\lambda^{X}*e^{-\lambda} }{X!}$

We get

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350\\$

The probability that at least 3 people arriving during a particular hour is 0.7350.

$EV=\lambda*t=5*0.75=3.75$

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2 years ago

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