3 questions, I'm horrible at this probability stuff!

A manufacturer makes two models of an item: model 1, which accounts for 80% of unit sales, and model 2, which accounts for 20% of all unit sales. Because of defects, the manufacturer has to replace (or exchange) 10% of its model 1 and 18% of its model 2. if a model is selected at random, find the probability that it will be defective.




Urn 1 contains 5 red balls and 3 black balls. Urn 2 contais 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If an urn is selected at random and a ball is drawn, find the probablity it will be red.




70.3% of females ages 20 t0 24 have never been married. Choose 5 young woman in this age category at random. Find the probability that a) none have ever been married and b) at least one has married.

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Originally Posted by aly111214

3 questions, I'm horrible at this probability stuff!

A manufacturer makes two models of an item: model 1, which accounts for 80% of unit sales, and model 2, which accounts for 20% of all unit sales. Because of defects, the manufacturer has to replace (or exchange) 10% of its model 1 and 18% of its model 2. if a model is selected at random, find the probability that it will be defective.

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=80% x 10% + 20% x 18% = 11.6%

Quote:

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Originally Posted by aly111214

Urn 1 contains 5 red balls and 3 black balls. Urn 2 contais 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. if an urn is selected at random and a ball is drawn, find the probablity it will be red.

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=1/3 x 5/8 + 1/3 x 3/4 + 1/3 x 4/6 = 68%

Quote:

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Originally Posted by aly111214

70.3% of females ages 20 t0 24 have never been married. Choose 5 young woman in this age category at random. Find the probability that a) none have ever been married and b) at least one has married.

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a) = (70.3%)^5 = 17%
b) = 1 - (70.3%)^5 = 83%

1. Let me explain:

The percentage of defects in model 1 is 10% of the 80%, which makes (10% x 80%) 8%.

The percentage of defects in model 2 is 18% of the 20%, which makes (18% x 20%) 3.6%.

Total percentage of defective models is (8% + 3.6%) 11.6%

Since the probability of having a defecting model is the amount of defective ones over the total models, then, the probability comes to 11.6% out of 100%, which is () 11.6%

2. The probability of having it from the 1st urn is 5/8 right?
The second one is 3/4
The third one is 4/6.

Ok, now probability of choosing the first urn is 1/3, as it is for urn 2 and 3.
So, probability of picking
a. Urn 1 and having a red ball is (1/3 x 5/8) 5/24.
b. That from urn 2 is (3/4 x 1/3) 1/4
c. That from urn 3 is (4/6 x 1/3) 2/9.

Now, it could be either of case a, b or c, so, you add the probabilities, (5/24 + 1/4 + 2/9) 49/72.

3.
a) Probability of having 5 unmarried females = probability 1st female unmarried, 2nd female unmarried, 3rd female unmarried, 4th female unmarried, 5th female unmarried.

Here you must have all unmarried to have the required probability, ans consecutively! That means you have to multiply the individual probabilities:

P(5 females unmarried) = P(unmarried) x P(unmarried) x P(unmarried) x P(unmarried) x P(unmarried) = (P(unmarried))^5 = 17.2%

b) Now, just remove that probability from 1. The only case when you do not have at least one unmarried female is when none are married. However, you just found it earlier. Use that answer to find this one.

I hope it helped! :)

Giving answers only not always is the good way to teach. The explanations are essential.