University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

196

Pages:

3
Author:

Beckett Gray

Question 1 A sample of 5 parts is drawn without replacement from a total population of 13 parts. Determine the probability of getting exactly 3 defective parts. The population is known to have 6 defective parts. Answer Step 1 Total Number of Population = 13 =13. Given, the population is known to have 6 6 defective parts. Probability of having a defective part: 𝑝 = 6 13 = 0.4615 p= 13 6 =0.4615 This distribution will follow a binomial distribution. Determine the probability of getting exactly 3 defective parts. Step 2 The full binomial probability formula with the binomial coefficient is: 𝑃 ( 𝑋 = 𝑥 ) = 𝑛 ! 𝑥 ! ( 𝑛 − 𝑥 ) ! ⋅ 𝑝 𝑥 ⋅ ( 1 − 𝑝 ) 𝑛 − 𝑥 P(X=x)= x!(n−x)! n! ⋅p x ⋅(1−p) n−x We need to compute 𝑃 ( 𝑋 = 3 ) P(X=3). This implies: 𝑃 ( 𝑋 = 3 ) = ( 5 3 ) ( 0.4615 ) 3 ( 0.5385 ) 5 − 3 P(X=3)=( 3 5 )(0.4615) 3 (0.5385) 5−3 = 5 ⋅ 4 ⋅ 3 3 ⋅ 2 ⋅ 1 ⋅ ( 0.4615 ) 3 ⋅ ( 0.5385 ) 2 = 3⋅2⋅1 5⋅4⋅3 ⋅(0.4615) 3 ⋅(0.5385) 2 = 10 ⋅ 0.0983 ⋅ 0.2909 =10⋅0.0983⋅0.2909 = 0.285 =0.285 Thus, the probability of getting exactly 3 defective parts is: 𝑃 ( 𝑋 = 3 ) = 0.285 P(X=3)=0.285 Question 2 Suppose 𝑋 ∼ 𝐵 ( 𝑛 = 10 , 𝑝 = 0.3 ) X∼B(n=10,p=0.3) using the binomial distribution table we have: 𝑃 ( 𝑋 < 4 ) P(X<4): a) 0.3828 b) 0.8497 c) 0.4059 d) 0.6496 Answer Step 1 Given 𝑝 = 0.3 p=0.3, so 1 − 𝑝 = 0.7 1−p=0.7. Step 2 Using the binomial probability distribution: 𝑃 ( 𝑋 < 4 ) = 𝑃 ( 𝑋 = 0 ) + 𝑃 ( 𝑋 = 1 ) + 𝑃 ( 𝑋 = 2 ) + 𝑃 ( 𝑋 = 3 ) P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3) = ( 10 0 ) ( 0.3 ) 0 ( 0.7 ) 10 + ( 10 1 ) ( 0.3 ) 1 ( 0.7 ) 9 + ( 10 2 ) ( 0.3 ) 2 ( 0.7 ) 8 + ( 10 3 ) ( 0.3 ) 3 ( 0.7 ) 7 =( 0 10 )(0.3) 0 (0.7) 10 +( 1 10 )(0.3) 1 (0.7) 9 +( 2 10 )(0.3) 2 (0.7) 8 +( 3 10 )(0.3) 3 (0.7) 7 = 0.0001 + 0.009 + 0.0459 + 0.1499 =0.0001+0.009+0.0459+0.1499 = 0.6496 =0.6496 Answer: 0.6496. Question 3 A fair die is rolled 12 times. What is the probability of rolling a 4 on the die 4 times? Answer We use the binomial probability formula: 𝑃 ( 𝑋 = 𝑥 ) = ( 𝑛 𝑥 ) 𝑝 𝑥 ( 1 − 𝑝 ) 𝑛 − 𝑥 P(X=x)=( x n )p x (1−p) n−x If a die is rolled, there are 6 outcomes ( 1 , 2 , 3 , 4 , 5 , 6 ) (1,2,3,4,5,6), and the probability of rolling a 4 is: 𝑝 = 1 6 p= 6 1 Here, 𝑛 = 12 n=12, 𝑥 = 4 x=4, and 𝑝 = 1 6 p= 6 1 . 𝑃 ( 𝑋 = 4 ) = ( 12 4 ) ( 1 6 ) 4 ( 5 6 ) 8 P(X=4)=( 4 12 )( 6 1 ) 4 ( 6 5 ) 8 = 12 ⋅ 11 ⋅ 10 ⋅ 9 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 1 1296 ⋅ 390625 1679616 = 4⋅3⋅2⋅1 12⋅11⋅10⋅9 ⋅ 1296 1 ⋅ 1679616 390625 = 495 ⋅ 0.000772 ⋅ 0.2326 =495⋅0.000772⋅0.2326 = 0.0888 =0.0888 Thus, the probability 𝑃 ( 𝑋 = 4 ) ≈ 0.0888 P(X=4)≈0.0888. Question 4 Simplify, please: 12 ! 8 ! 4 ! 8!4! 12! . Answer 1 12 ! 8 ! 4 ! = 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ! 8 ! ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 8!4! 12! = 8!⋅4⋅3⋅2⋅1 12⋅11⋅10⋅9⋅8! = 12 ⋅ 11 ⋅ 10 ⋅ 9 4 ⋅ 3 ⋅ 2 ⋅ 1 = 4⋅3⋅2⋅1 12⋅11⋅10⋅9 = 11880 24 = 495 = 24 11880 =495 Answer 2 12 ! 8 ! 4 ! = 495 8!4! 12! =495 Question 5 What does it mean 5 𝐶 3 5C3? Solve, please. Answer 1 5 𝐶 3 = 5 ! 3 ! ( 5 − 3 ) ! 5C3= 3!(5−3)! 5! = 5 ⋅ 4 ⋅ 3 ! 3 ! ⋅ 2 ⋅ 1 = 3!⋅2⋅1 5⋅4⋅3! = 5 ⋅ 4 2 ⋅ 1 = 20 2 = 10 = 2⋅1 5⋅4 = 2 20 =10 Answer 2 The general formula is: 𝑛 𝐶 𝑟 = 𝑛 ! 𝑟 ! ( 𝑛 − 𝑟 ) ! nC r = r!(n−r)! n! For 𝑛 = 5 , 𝑟 = 3 n=5,r=3: 5 𝐶 3 = 10 5C3=10