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wariber [46]
3 years ago
11

2. The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability th

at he must stop at the first signal is 0.36, the analogous probability for the second signal is 0.54, and the probability that he must stop at at least one of the two signals is 0.65. What is the probability that he must stop at exactly one signal?
Mathematics
1 answer:
coldgirl [10]3 years ago
8 0

Answer:

0.30

Step-by-step explanation:

Probability of stopping at first signal = 0.36 ;

P(stop 1) = P(x) = 0.36

Probability of stopping at second signal = 0.54;

P(stop 2) = P(y) = 0.54

Probability of stopping at atleast one of the two signals:

P(x U y) = 0.6

Stopping at both signals :

P(xny) = p(x) + p(y) - p(xUy)

P(xny) = 0.36 + 0.54 - 0.6

P(xny) = 0.3

Stopping at x but not y

P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06

Stopping at y but not x

P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24

Probability of stopping at exactly 1 signal :

P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30

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Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

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Which graph shows a line with positive slope that passes through the point (r, s)?
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So far the only true statement I see is that function f is increasing whilst function g is decreasing.

I cannot say yes to the first one as there is a lack of x-intercepts.

Past the interval of 0,2, there are no changes in the line of function g as it remains the same despite the increase in y. Function f on the other hand has a rate of change. 

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Answer:

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2 kg = 2000g

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600 : 2000

divide both sides by a common divisor ( 200)

600/200 : 2000/200

3 : 10

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