Binomial probability #9 - Questions and Answers

Question 1 Plastic drinking straws are manufactured to fit into drinks cartons which have a hole in the top. A straw fits into the hole if the diameter of the straw is less than 3 mm. The diameters of the straws have a normal distribution with mean 2.6 mm and standard deviation 0.25 mm. (a) A straw is chosen at random. Find the probability that it fits into the hole in a drinks carton. (b) 500 straws are chosen at random. Use a suitable approximation to find the probability that at least 480 straws fit into the holes in drinks cartons. Answer 1 Step 1 (a) If the diameter of the straw is less than 3 mm, it can fit in the hole. The standard deviation is 0.25 mm, and the mean is 2.6 mm. The likelihood that it will fit in the hole in a beverage carton is, 𝑃 ( 𝑋 < 3 ) = 𝑃 ( 𝑋 − 𝜇 𝜎 < 3 − 2.6 0.25 ) P(X<3)=P( σ X−μ ​ < 0.25 3−2.6 ​ ) = 𝑃 ( 0.4 0.25 ) =P( 0.25 0.4 ​ ) = 𝑃 ( 𝑍 < 1.60 ) =P(Z<1.60) The probability of 𝑍 Z less than 1.60 can be obtained using the excel formula "=NORM.S.DIST(1.60, TRUE)." The probability value is 0.9452. The required probability is: 𝑃 ( 𝑋 < 3 ) = 𝑃 ( 𝑍 < 1.60 ) = 0.9452 P(X<3)=P(Z<1.60)=0.9452 Thus, the probability that a straw fits into the hole in a drinks carton is 0.9452. Step 2 (b) Let 𝑋 X be the number of straws that fit into the holes in drinks cartons. This follows a binomial distribution with sample size 𝑛 = 500 n=500 and probability of success 𝑝 = 0.9452 p=0.9452. The mean is: 𝜇 = 𝑛 𝑝 = 500 ⋅ 0.9452 = 472.6 μ=np=500⋅0.9452=472.6 The standard deviation is: 𝜎 = 𝑛 𝑝 ( 1 − 𝑝 ) = 500 ⋅ 0.9452 ⋅ ( 1 − 0.9452 ) = 4.6745 σ= np(1−p) ​ = 500⋅0.9452⋅(1−0.9452) ​ =4.6745 Using a binomial model with a suitable normal approximation, apply the continuity correction: 𝑃 ( 𝑋 ≥ 480 ) = 𝑃 ( 𝑋 > 479.5 ) P(X≥480)=P(X>479.5) = 1 − 𝑃 ( 𝑋 − 𝜇 𝜎 < 479.5 − 472.6 4.6745 ) =1−P( σ X−μ ​ < 4.6745 479.5−472.6 ​ ) = 1 − 𝑃 ( 𝑍 < 0.5134 ) =1−P(Z<0.5134) The probability of 𝑍 Z less than 0.5134 can be obtained using the excel formula "=NORM.S.DIST(0.5134, TRUE)." The probability value is 0.6962. The required probability is: 𝑃 ( 𝑋 ≥ 480 ) = 1 − 𝑃 ( 𝑍 < 0.5134 ) = 1 − 0.6962 = 0.3038 P(X≥480)=1−P(Z<0.5134)=1−0.6962=0.3038 Hence, the probability that at least 480 straws fit into the holes in drinks cartons is 0.3038. Answer 2 Step 1 𝑋 ∼ 𝐵 ( 500 , 0.9452 ) ∼ 𝑁 ( 472.6 , 25.898 ) X∼B(500,0.9452)∼N(472.6,25.898) Standardizing: 𝑍 = 479.5 − 472.6 500 ⋅ 0.9452 ⋅ ( 1 − 0.9452 ) Z= 500⋅0.9452⋅(1−0.9452) ​ 479.5−472.6 ​ 𝑍 = 6.9 4.6745 ≈ 1.48 Z= 4.6745 6.9 ​ ≈1.48 Using standard normal tables, 𝑃 ( 𝑍 > 1.48 ) = 0.0694 P(Z>1.48)=0.0694. Thus, the probability 𝑃 ( 𝑋 ≥ 480 ) P(X≥480) is approximately 0.3038. Question 2 The probability that a student passes the Probability and Statistics exam is 0.7. (i) Find the probability that a given student will pass the Probability and Statistics exam after the third try. (ii) Find the probability that the third student who fails the Probability and Statistics exam is found after the fifth student in a class of 9 students. (iii) Find the probability that at most 3 students will pass the Probability and Statistics exam in a class of 7 students. Answer 1 Step 1 The probability of passing the exam is 𝑝 = 0.7 p=0.7, so the probability of failing is 𝑞 = 1 − 𝑝 = 0.3 q=1−p=0.3. (i) The student passes after the third try, which means they fail twice and pass on the third attempt. The probability is: 𝑃 = 𝑞 ⋅ 𝑞 ⋅ 𝑝 = ( 0.3 ) ⋅ ( 0.3 ) ⋅ ( 0.7 ) = 0.063 P=q⋅q⋅p=(0.3)⋅(0.3)⋅(0.7)=0.063 Step 2 (ii) The third student who fails is found after the fifth student, meaning 2 students failed in the first 5 trials and 1 student failed in the remaining 4 trials. Use the binomial distribution formula: 𝑃 = ( 5 2 ) ( 0.3 ) 2 ( 0.7 ) 3 ⋅ ( 4 1 ) ( 0.3 ) 1 ( 0.7 ) 3 P=( 2 5 ​ )(0.3) 2 (0.7) 3 ⋅( 1 4 ​ )(0.3) 1 (0.7) 3 𝑃 ≈ 0.127 P≈0.127 Step 3 (iii) To find the probability that at most 3 students pass the exam among 7 students, sum the probabilities for 𝑋 = 0 , 1 , 2 , 3 X=0,1,2,3: 𝑃 ( 𝑋 ≤ 3 ) = 𝑃 ( 𝑋 = 0 ) + 𝑃 ( 𝑋 = 1 ) + 𝑃 ( 𝑋 = 2 ) + 𝑃 ( 𝑋 = 3 ) P(X≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) Using the binomial probability formula: 𝑃 ( 𝑋 = 𝑘 ) = ( 7 𝑘 ) ( 0.7 ) 𝑘 ( 0.3 ) 7 − 𝑘 P(X=k)=( k 7 ​ )(0.7) k (0.3) 7−k 𝑃 ( 𝑋 ≤ 3 ) ≈ 0.126 P(X≤3)≈0.126 Question 3 If a baseball player has a batting average of 0.275, what is the likelihood that the player will get the following number of hits in the next four times at bat? (A) Exactly 2 hits (B) At least 2 hits Answer 1 Step 1 The probability of success is 𝑝 = 0.275 p=0.275 and failure is 𝑞 = 1 − 𝑝 = 0.725 q=1−p=0.725. Use the binomial distribution for 𝑛 = 4 n=4: 𝑃 ( 𝑋 = 𝑥 ) = ( 4 𝑥 ) ( 0.275 ) 𝑥 ( 0.725 ) 4 − 𝑥 P(X=x)=( x 4 ​ )(0.275) x (0.725) 4−x Step 2 (A) Exactly 2 hits: 𝑃 ( 𝑋 = 2 ) = ( 4 2 ) ( 0.275 ) 2 ( 0.725 ) 2 P(X=2)=( 2 4 ​ )(0.275) 2 (0.725) 2 𝑃 ( 𝑋 = 2 ) ≈ 0.239 P(X=2)≈0.239 (B) At least 2 hits: 𝑃 ( 𝑋 ≥ 2 ) = 𝑃 ( 𝑋 = 2 ) + 𝑃 ( 𝑋 = 3 ) + 𝑃 ( 𝑋 = 4 ) P(X≥2)=P(X=2)+P(X=3)+P(X=4) 𝑃 ( 𝑋 ≥ 2 ) ≈ 0.305 P(X≥2)≈0.305