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cricket20 [7]
3 years ago
10

A box contains five red balls, six green ones, and nine yellow balls. Suppose you select one ball at random from the box and do

not replace it. Then you randomly select a second ball. Find the probability that both balls will be red?
Mathematics
1 answer:
Tanya [424]3 years ago
8 0

Answer:1/19

Step-by-step explanation:

Number of red balls: 5

Number of green balls: 6

Number of yellow balls: 9

Total number of balls= 5+6+9= 20

Let probability of red balls = Pr(red balls)

PR(red balls)= 5/20

Hence probability of picking first ball as red = 5/20

Probability of picking second ball without replacing the first ball:

This implies that by picking second ball to be red, the number of red ball would decrease by 1. Also the total number of balls would also decrease by 1 because it isn't replaced.

PR(2nd ball) = 4/19

Pr(1st ball red and 2nd ball red)= 5/20 * 4/19

=(5*4)/(20×19)

20/(20/19)

= 1/19

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For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger
11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

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The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419

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