Question

In a group of

10

​batteries,

5

are dead. You choose 2 batteries at random.

​a) Create a probability model for the number of good batteries you get.

​b) What's the expected number of good ones you​ get?

​c) What's the standard​ deviation?

​a) Create a probability model.

Number good	0	1	2
​P(Number good)	nothing	nothing	nothing

​(Round to three decimal places as​ needed.)

​b) The expected number of good batteries is

nothing.

​(Round to two decimal places as​ needed.)

​c) The standard deviation is

nothing.

Answer 1

a)

Probability of not getting a good battery = ⁵C₂ / ¹⁰C₂ = 10/45 = 2/9

Probability of getting a good battery and a bad battery = ⁵C₁ * ⁵C₁ / ¹⁰C₂ = 25/45 = 5/9

Probability of not getting both good battery = ⁵C₂ / ¹⁰C₂ = 10/45 = 2/9

Number (Good)	0	1	2
P(Number Good)	2/9	5/9	2/9

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b) The Expected number of good batteries = SUM[ x * P(x)]

= 0 * (2/9) + 1 * (5/9) + 2 * (2/9)

= 0 + 5/9 + 4/9

= 9/9

= 1.00

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c) Standard Deviation = Sqrt(Variance)

Variance = E(x²) - [E(x)]²

E(x²) = SUM[x² * P(x)] = (0² * (2/9)) + (1² * (5/9)) + (2² * (2/9))

= 0 + 5/9 + 8/9

= 13/9

[E(x)]² = (1)² = 1

Therefore Variance = 13/9 - 1 = 4/9

Standard Deviation = Sqrt(Variance) = Sqrt(4/9) = 2/3 = 0.67

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