Question

find the probability, to 4 decimal places: it is possible when rounded that a probability is 0.0000 d) exactly none are left - handed. e) exactly 10 are left - handed. f) at least 9 are left - handed. g) at most 10 are left - handed. h) at least 12 are left - handed.

Explanation:

1. Assume a binomial - distribution:

• Let's assume that the number of left - handed people \(X\) follows a binomial distribution \(X\sim B(n,p)\), where \(n\) is the total number of people in the sample and \(p\) is the probability of an individual being left - handed. However, since \(n\) and \(p\) are not given in the problem, for the sake of generality, the binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).
• d) Probability that exactly none are left - handed (\(k = 0\)):
• \(P(X = 0)=C(n,0)\times p^{0}\times(1 - p)^{n-0}=(1 - p)^{n}\). Without knowing \(n\) and \(p\), if we assume \(p = 0.1\) (a common estimate for the proportion of left - handed people in the general population) and say \(n = 20\) (arbitrary sample size for illustration).
• \(C(20,0)=\frac{20!}{0!(20 - 0)!}=1\), \(p^{0}=1\). Then \(P(X = 0)=(1 - 0.1)^{20}\).
• # Explanation:

Step1: Calculate the binomial coefficient

\(C(20,0)=\frac{20!}{0!×20!}=1\)

Step2: Substitute into the binomial formula

\(P(X = 0)=1\times1\times(0.9)^{20}\approx0.1216\)

• e) Probability that exactly 10 are left - handed (\(k = 10\)):
• \(P(X = 10)=C(n,10)\times p^{10}\times(1 - p)^{n - 10}\). Using \(n = 20\) and \(p = 0.1\).
• \(C(20,10)=\frac{20!}{10!(20 - 10)!}=\frac{20!}{10!×10!}=\frac{20\times19\times\cdots\times11}{10!}=184756\).
• \(P(X = 10)=184756\times(0.1)^{10}\times(0.9)^{10}\).
• # Explanation:

Step1: Calculate the binomial coefficient

\(C(20,10)=\frac{20!}{10!×10!}=184756\)

Step2: Substitute into the binomial formula

\(P(X = 10)=184756\times(0.1)^{10}\times(0.9)^{10}\approx0.0000\)

• f) Probability that at least 9 are left - handed (\(P(X\geq9)=P(X = 9)+P(X = 10)+\cdots+P(X = n)\)):
• \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\). For \(n = 20\) and \(p = 0.1\).
• \(P(X = 9)=C(20,9)\times(0.1)^{9}\times(0.9)^{11}\), \(C(20,9)=\frac{20!}{9!(20 - 9)!}=\frac{20!}{9!×11!}=167960\).
• \(P(X = 9)=167960\times(0.1)^{9}\times(0.9)^{11}\approx0.0000\).
• \(P(X\geq9)=\sum_{k = 9}^{20}C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\approx0.0000\)
• # Explanation:

Step1: Calculate individual binomial probabilities

For \(k\) from 9 to 20, calculate \(C(20,k)=\frac{20!}{k!(20 - k)!}\) and \(P(X = k)=C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\)

Step2: Sum the probabilities

\(P(X\geq9)=\sum_{k = 9}^{20}P(X = k)\approx0.0000\)

• g) Probability that at most 10 are left - handed (\(P(X\leq10)=\sum_{k = 0}^{10}P(X = k)=\sum_{k = 0}^{10}C(n,k)\times p^{k}\times(1 - p)^{n - k}\)):
• Using \(n = 20\) and \(p = 0.1\).
• \(P(X\leq10)=\sum_{k = 0}^{10}C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\).
• We know \(P(X = 0)=(0.9)^{20}\), \(P(X = 1)=C(20,1)\times(0.1)^{1}\times(0.9)^{19}=20\times0.1\times(0.9)^{19}\), etc.
• \(P(X\leq10)\approx1.0000\)
• # Explanation:

Step1: Calculate individual binomial probabilities

For \(k\) from 0 to 10, calculate \(C(20,k)=\frac{20!}{k!(20 - k)!}\) and \(P(X = k)=C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\)

Step2: Sum the probabilities

\(P(X\leq10)=\sum_{k = 0}^{10}P(X = k)\approx1.0000\)

• h) Probability that at least 12 are left - handed (\(P(X\geq12)=P(X = 12)+P(X = 13)+\cdots+P(X = n)\)):
• Using \(n = 20\) and \(p = 0.1\).
• \(P(X\geq12)=\sum_{k = 12}^{20}C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\approx0.0000\)
• #…

Answer:

1. Assume a binomial - distribution:

• Let's assume that the number of left - handed people \(X\) follows a binomial distribution \(X\sim B(n,p)\), where \(n\) is the total number of people in the sample and \(p\) is the probability of an individual being left - handed. However, since \(n\) and \(p\) are not given in the problem, for the sake of generality, the binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).
• d) Probability that exactly none are left - handed (\(k = 0\)):
• \(P(X = 0)=C(n,0)\times p^{0}\times(1 - p)^{n-0}=(1 - p)^{n}\). Without knowing \(n\) and \(p\), if we assume \(p = 0.1\) (a common estimate for the proportion of left - handed people in the general population) and say \(n = 20\) (arbitrary sample size for illustration).
• \(C(20,0)=\frac{20!}{0!(20 - 0)!}=1\), \(p^{0}=1\). Then \(P(X = 0)=(1 - 0.1)^{20}\).
• # Explanation:

Step1: Calculate the binomial coefficient

\(C(20,0)=\frac{20!}{0!×20!}=1\)

Step2: Substitute into the binomial formula

\(P(X = 0)=1\times1\times(0.9)^{20}\approx0.1216\)

• e) Probability that exactly 10 are left - handed (\(k = 10\)):
• \(P(X = 10)=C(n,10)\times p^{10}\times(1 - p)^{n - 10}\). Using \(n = 20\) and \(p = 0.1\).
• \(C(20,10)=\frac{20!}{10!(20 - 10)!}=\frac{20!}{10!×10!}=\frac{20\times19\times\cdots\times11}{10!}=184756\).
• \(P(X = 10)=184756\times(0.1)^{10}\times(0.9)^{10}\).
• # Explanation:

Step1: Calculate the binomial coefficient

\(C(20,10)=\frac{20!}{10!×10!}=184756\)

Step2: Substitute into the binomial formula

\(P(X = 10)=184756\times(0.1)^{10}\times(0.9)^{10}\approx0.0000\)

• f) Probability that at least 9 are left - handed (\(P(X\geq9)=P(X = 9)+P(X = 10)+\cdots+P(X = n)\)):
• \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\). For \(n = 20\) and \(p = 0.1\).
• \(P(X = 9)=C(20,9)\times(0.1)^{9}\times(0.9)^{11}\), \(C(20,9)=\frac{20!}{9!(20 - 9)!}=\frac{20!}{9!×11!}=167960\).
• \(P(X = 9)=167960\times(0.1)^{9}\times(0.9)^{11}\approx0.0000\).
• \(P(X\geq9)=\sum_{k = 9}^{20}C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\approx0.0000\)
• # Explanation:

Step1: Calculate individual binomial probabilities

For \(k\) from 9 to 20, calculate \(C(20,k)=\frac{20!}{k!(20 - k)!}\) and \(P(X = k)=C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\)

Step2: Sum the probabilities

\(P(X\geq9)=\sum_{k = 9}^{20}P(X = k)\approx0.0000\)

• g) Probability that at most 10 are left - handed (\(P(X\leq10)=\sum_{k = 0}^{10}P(X = k)=\sum_{k = 0}^{10}C(n,k)\times p^{k}\times(1 - p)^{n - k}\)):
• Using \(n = 20\) and \(p = 0.1\).
• \(P(X\leq10)=\sum_{k = 0}^{10}C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\).
• We know \(P(X = 0)=(0.9)^{20}\), \(P(X = 1)=C(20,1)\times(0.1)^{1}\times(0.9)^{19}=20\times0.1\times(0.9)^{19}\), etc.
• \(P(X\leq10)\approx1.0000\)
• # Explanation:

Step1: Calculate individual binomial probabilities

For \(k\) from 0 to 10, calculate \(C(20,k)=\frac{20!}{k!(20 - k)!}\) and \(P(X = k)=C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\)

Step2: Sum the probabilities

\(P(X\leq10)=\sum_{k = 0}^{10}P(X = k)\approx1.0000\)

• h) Probability that at least 12 are left - handed (\(P(X\geq12)=P(X = 12)+P(X = 13)+\cdots+P(X = n)\)):
• Using \(n = 20\) and \(p = 0.1\).
• \(P(X\geq12)=\sum_{k = 12}^{20}C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\approx0.0000\)
• # Explanation:

Step1: Calculate individual binomial probabilities

For \(k\) from 12 to 20, calculate \(C(20,k)=\frac{20!}{k!(20 - k)!}\) and \(P(X = k)=C(20,k)\times(0.1)^{k}\times(0.9)^{20 - k}\)

Step2: Sum the probabilities

\(P(X\geq12)=\sum_{k = 12}^{20}P(X = k)\approx0.0000\)

1. Answer:

• d) 0.1216
• e) 0.0000
• f) 0.0000
• g) 1.0000
• h) 0.0000