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VLD [36.1K]
3 years ago
5

A bowl contains 25 balls numbered 1 to 25. A ball is drawn and its number is noted. Without replacing the first ball, another ba

ll is drawn.
The probability that the numbers on both balls are odd numbers is .
Mathematics
1 answer:
mamaluj [8]3 years ago
6 0
Let A = the event that the first ball is odd number
and B = the event that the second ball is odd number

In the beginning, there are 25 balls which contain 13 balls of odd numbers. So the probability for the first selection (as to expect you would draw a ball of an odd number) is
P(A) = 13/25

After the selection, the rule does not allow replacement. So, there are 24 balls left containing now 12 balls of odd numbers. So the probability for the second selection (as to expect you would draw also a ball of odd number) is
P(B) = 12/24 = 1/2

Multiplying the two probabilities obtains
P(2 odd numbers) = 13/25 * 1/2 = 13/50

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Line passing through (2,5) and the m= -5/6
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Answer:

y - 5 = (-5/6)(x - 2)

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Step-by-step explanation:

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3 years ago
Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f
castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean (\mu) = 115 \times 20

= 2300

Standard deviation = 20 \times \sqrt{115}

Standard deviation (SD) = 214.4761

TO find:

a) P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})

P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})

P(x > 3500)= P(Z > \dfrac{1200}{214.4761})

P(x > 3500)= P(Z >5.595)

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P(x > 3500)=1-0.9999

P(x > 3500) = 0.0001

b)

Here, the replacement time for the mean (\mu) = \dfrac{0+0.5}{2}

= 0.25

Replacement time for the Standard deviation \sigma = \dfrac{0.5-0}{\sqrt{12}}

\sigma = 0.1443

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = \sqrt{(115\times 20^2) +(114\times (0.1443)^2)}

= \sqrt{(115\times 400) +(114\times 0.02082249}

= \sqrt{(46000) +2.37376386}

= \sqrt{(46000) +(2.37376386)}

= \sqrt{46002.374}

= 214.482

Now; the required probability:

P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})

P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})

P(x > 4125) = P(Z >8.376)

P(x > 4125) =1-  P(Z

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

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3 years ago
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