A bag contains a total of 20 batteries, of which four are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good.
1. 3/95 - My answer
2. 3/5
3. 1/5
4. 4/95
8. Evaluate: 11C6
1. 66
2. 120
3. 24
4. 462 - My answer
9. Given that P(A) = 0.5, P(B) = 0.4, and P(A and B) = 0.20, determine P(A|B)
1. 0.40 - My answer
2. 0.50
3. 0.25
4. 0.20
10. If P(A or B) = 0.8, P(A) = 0.8 and P(B) = 0.5, determine P(A and B).
1. 0.35
2. 0.50 - My answer
3. 0.64
4. 0.13
Also, could someone help me with this, please?
Given the bar graph below, create a grouped frequency distribution
The graph: http://www.flickr.com/photos/62474130@N0…
1. 3/95 - My answer
2. 3/5
3. 1/5
4. 4/95
8. Evaluate: 11C6
1. 66
2. 120
3. 24
4. 462 - My answer
9. Given that P(A) = 0.5, P(B) = 0.4, and P(A and B) = 0.20, determine P(A|B)
1. 0.40 - My answer
2. 0.50
3. 0.25
4. 0.20
10. If P(A or B) = 0.8, P(A) = 0.8 and P(B) = 0.5, determine P(A and B).
1. 0.35
2. 0.50 - My answer
3. 0.64
4. 0.13
Also, could someone help me with this, please?
Given the bar graph below, create a grouped frequency distribution
The graph: http://www.flickr.com/photos/62474130@N0…
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I agree with #1, #8, and #10
#9: P(A|B) = P(A and B)/P(B)
0.2/0.4 = 0.5 http://www.mathgoodies.com/lessons/vol6/…
Your teacher may want the grouped frequency distribution in alphabetical order?
Basically, all your going to do is make a table:
Country Hours
Sweeden 6.9
France 6.9
Russia 7.1
etc..
#9: P(A|B) = P(A and B)/P(B)
0.2/0.4 = 0.5 http://www.mathgoodies.com/lessons/vol6/…
Your teacher may want the grouped frequency distribution in alphabetical order?
Basically, all your going to do is make a table:
Country Hours
Sweeden 6.9
France 6.9
Russia 7.1
etc..
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all ok !
cheers !
cheers !
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all are right xD