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Mekhanik [1.2K]
3 years ago
15

suppose x is a non-negative integer valued random variable which follows a probability distribution such that p x=0 = 06 and for

each i>=1, P(X=i+1) = 1/5 P(x=i). Calculate P(X<=2)
Mathematics
1 answer:
Andre45 [30]3 years ago
6 0

Not sure if the given probability that X=0 is 0.06 or 0.6, or something else altogether. I'll just refer to it by the number p.

\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge1\end{cases}

As is, the probability that X=1 is indeterminate, so I think you intended to write

\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge\boxed{0}\end{cases}

Then by this definition,

i=0\implies P(X=1)=\dfrac15P(X=0)=\dfrac p5

i=1\implies P(X=2)=\dfrac15P(X=1)=\dfrac p{5^2}

and so on.

Then

\boxed{P(X\le2)=P(X=0)+P(X=1)+P(X=2)=p+\dfrac p5+\dfrac p{5^2}=\dfrac{31p}{25}}

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Answer:

y=4(3)^x

Step-by-step explanation:

From the table given in the question,

x            y              Difference                  Ratio

1            12                     -                             -

2           36              36-12 = 24              \frac{36}{12}=3

3          108            108-36 = 72             \frac{108}{36}=3

4          324            324-108 = 216        \frac{324}{108}=3

5          972            972-324 = 648      \frac{648}{216}=3

There is a common ratio of 3 in each successive term.

Therefore, data given in the table will represent an exponential function.

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For a point (1, 12),

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For another point of the table (2,36),

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Equation (2) divided by equation (1)

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