Thread: The Binomial Distribution errors?

just thought i would go on a few sites and pratice some Binomial Distribution questions
i found these on tutorvista , is it me or are all the answers given wrong



Examples for Application of Binomial Disrtibution:
Example 1:
A die is tossed 4 times. What is the Probability of getting exactly 3 fours?
Solution:
Here n = 4, x = 3, probability of success on a single trial = 1/ 4 or 0.25.
Therefore, The binomial probability is,
p(X =r) = ncr pr q(n-r) where p + q=1 then q =1-p
q=1-0.25
q=0.75
b( 3; 4, 0.25 ) = 4C3 × ( 0.25)3 × ( 0.75)4 −3
= ( 4! / 3! × (4-3)!) × 0.016 × ( 0.75)
= (4! / 3! × 1!) × 0.016× 0.75
= 4 × 0.016 × 0.75
b( 3; 4, 0.25 ) = 0.048. Answer.
Example 2:
A die is tossed 7 times. What is the Probability of getting exactly 5 twos?
Solution:
Here n = 7, x = 5, probability of success on a single trial = 1/ 7 or 0.143.
Therefore, The binomial probability is,
p(X =r) = ncr pr q(n-r) where p + q=1 then q =1-p
q=1-0.143
q=0.857
b( 5; 7, 0.143 ) = 7C5 × ( 0.143)5 × ( 0.857)7 −5
= ( 7! / 5! × (7-5)!) × (0.0001) × ( 0.857)2
= (7! / 5! × 2!) × 0.0001× 0.734
= 21 ×0.0001 × 0.734
b( 5; 7, 0.143 ) = 0.00154. Answer.

Practice Problems for Application of Binomial Distribution:

1 A die is tossed 2 times. What is the Probability of getting exactly 1sixes?

1) 0.5.

Originally Posted by jickjoker

just thought i would go on a few sites and pratice some Binomial Distribution questions
i found these on tutorvista , is it me or are all the answers given wrong



Examples for Application of Binomial Disrtibution:
Example 1:
A die is tossed 4 times. What is the Probability of getting exactly 3 fours?
Solution:
Here n = 4, x = 3, probability of success on a single trial = 1/ 4 or 0.25.
Therefore, The binomial probability is,
p(X =r) = ncr pr q(n-r) where p + q=1 then q =1-p
q=1-0.25
q=0.75
b( 3; 4, 0.25 ) = 4C3 × ( 0.25)3 × ( 0.75)4 −3
= ( 4! / 3! × (4-3)!) × 0.016 × ( 0.75)
= (4! / 3! × 1!) × 0.016× 0.75
= 4 × 0.016 × 0.75
b( 3; 4, 0.25 ) = 0.048. Answer.
Example 2:
A die is tossed 7 times. What is the Probability of getting exactly 5 twos?
Solution:
Here n = 7, x = 5, probability of success on a single trial = 1/ 7 or 0.143.
Therefore, The binomial probability is,
p(X =r) = ncr pr q(n-r) where p + q=1 then q =1-p
q=1-0.143
q=0.857
b( 5; 7, 0.143 ) = 7C5 × ( 0.143)5 × ( 0.857)7 −5
= ( 7! / 5! × (7-5)!) × (0.0001) × ( 0.857)2
= (7! / 5! × 2!) × 0.0001× 0.734
= 21 ×0.0001 × 0.734
b( 5; 7, 0.143 ) = 0.00154. Answer.

Practice Problems for Application of Binomial Distribution:

1 A die is tossed 2 times. What is the Probability of getting exactly 1sixes?

1) 0.5.

Edit:

In your first example you have p = 1/4. Is that what the solution said or what the questions said. If the die is fair, then p = 1/6 not 1/4. Similarly on example 2, for a fair die p = 1/6 not 1/7. It looks to me like you should be referencing more competent websites ....

Practice question: Assuming a fair die, I get 0.2778 (correct to 4dp). Show your working if you want help with why you're wrong.

example 1 .
A die is tossed 4 times. What is the Probability of getting exactly 3 fours?
4c3 = 4 * 0.166 * 0.166* 0.166* 0.833 = 0.01524

example 2 .
A die is tossed 7 times. What is the Probability of getting exactly 5 twos?
7c5 = 21 * 0.166 * 0.166* 0.166 *0.166* 0.166* 0.833 * 0.833 =0.001830

Practice question:
2 * 0.166 * 0.833 = 0.2776