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Tema [17]
3 years ago
8

Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls

numbered 1, . . . , 12. Finally, box C contains four balls numbered 1, . . . , 4. One ball is selected from each urn uniformly at random.
(a) What is the probability that the ball chosen from box A is labeled 1 if exactly two balls numbered 1 were selected
(b) What is the probability that the ball chosen from box B is 12 if the arithmetic mean of the three balls selected is exactly 7?
Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222

Hence,

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684

Part b

P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}

Hence,

P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}

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A pole that is 3.4 m tall casts a shadow that is 1.54 m long. At the same time, a nearby building casts a shadow that is 35.75 m
inna [77]

Answer:

The building would be 78.92 meters tall.

Step-by-step explanation:

If the shodow of a 3.4m tall pole is 1.54m tall, and you have the shadow's height, then you make a cross, multiply-divide ratio table.

5 0
2 years ago
1. A baseball player got a hit 19 times in his last 64 times at bat.
fiasKO [112]

A) Probability =0.297

B)In 200 times he can hit 59 times !

<u>Step-by-step explanation:</u>

Here we have , A baseball player got a hit 19 times in his last 64 times at bat. We need to find the following :

a. What is the experimental probability that the player gets a hit in an at bat?

According to question ,

Favorable outcomes = 19

Total outcomes = 64

Probability = (Favorable outcomes)/(Total outcomes) i.e.

⇒ Probability = \frac{19}{64}

⇒ Probability =0.297

b. If the player comes up to bat 200 times in a season, about how many hits is he likely to get?​

According to question , In 64 times he hit 19 times . In 1 time there's probability to hit 0.297 times! So ,In 200 times he can hit :

⇒ Hit =0.297(200)

⇒ Hit = 59.36

Therefore , In 200 times he can hit 59 times !

4 0
3 years ago
Need help asap pls pls pls
Sliva [168]

Answer:

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Step-by-step explanation:

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6 0
2 years ago
Red Bull is the most popular energy drink in sales in the United States. Red Bull GmbH (the parent company) has observed that da
oee [108]

Answer:

The probability that on a given day below 6,214,323 drinks are sold is 0.8413.

Step-by-step explanation:

The provided information are:

Population mean (\mu) = 6,205,195

Population standard deviation (\sigma) = 9,120.32

Consider, <em>X</em> be the random variable that represents the number of drinks that are sold on a given that is normally distributed with the mean = 6,205,195 and standard deviation = 9,120.32.

The probability that on a given day below 6,214,323 drinks are sold can be calculated as:

P(X

It must be noted that <em>P</em>(<em>Z</em> < 1.0008) = 0.8413 has been calculated using the standard normal table.

Hence, the required probability is 0.8413.

8 0
3 years ago
Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp
aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}

Using a statistic software I've calculated the test

F_{H_0}= 1.41

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03

t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

8 0
3 years ago
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