Question

The National Hockey League (NHL) has 80% of its players born outside the United States, and of those born outside the United States, 50% are born in Canada. Suppose that n=12 NHL players were selected at random. Let x be the number of players in the sample who were born outside of the United States so that p=.8. Find the following probabilities: a. At least five or more of the sample players were born outside the United States. b. Exactly seven of the players were born outside the United States. c. Fewer than six were born outside the United States.

Answer 1

 

DEFINITIONS

Definition $\text{binomial probability}$:

$$P(X=k)=C^k\cdot p^k\cdot q^{n-k}=\genfrac{}{}{0ex}{}{n!}{k!(n-k)!}{p}^k(1-p{)}^{n-k}$$

$\text{Addition rule}$ for disjoint or mutually exclusive events:

$$P(A\cup B)=P(A\text{ or }B)=P(A)+P(B)$$

$\text{Complement rule}$:

$$P(A^c)=P(\text{ not }A)=1-P(A)$$

SOLUTION

$$\begin{array}{rl}n& =\text{Number of trials}=12\\ p& =\text{Probability of success}=80\%=0.8\end{array}$$

(a) Evaluate the definition of binomial probability at $k=0,1,2,3,4$:

$$\begin{array}{rl}P(X=0)& =C^0\cdot 0.7^0\cdot (1-0.7{)}^{12-0}=\genfrac{}{}{0ex}{}{12!}{0!(12-0)!}\cdot 0.7^0\cdot 0.7^{12}\approx 0.0000\\ P(X=1)& =C^1\cdot 0.7^1\cdot (1-0.7{)}^{12-1}=\genfrac{}{}{0ex}{}{12!}{1!(12-1)!}\cdot 0.7^1\cdot 0.7^{11}\approx 0.0000\\ P(X=2)& =C^2\cdot 0.7^2\cdot (1-0.7{)}^{12-2}=\genfrac{}{}{0ex}{}{12!}{2!(12-2)!}\cdot 0.7^2\cdot 0.7^{10}\approx 0.0000\\ P(X=3)& =C^3\cdot 0.7^3\cdot (1-0.7{)}^{12-3}=\genfrac{}{}{0ex}{}{12!}{3!(12-3)!}\cdot 0.7^3\cdot 0.7^9\approx 0.0001\\ P(X=4)& =C^4\cdot 0.7^4\cdot (1-0.7{)}^{12-4}=\genfrac{}{}{0ex}{}{12!}{4!(12-4)!}\cdot 0.7^4\cdot 0.7^8\approx 0.0005\end{array}$$

Since it is not possible to obtain two different number of successes on the same simulation, we can use the addition rule for mutually exclusive events:

$$\begin{array}{rl}P(X\le 4)& =P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\\ & =0.0000+0.0000+0.0000+0.0001+0.0005\\ & =0.0006\end{array}$$

Use the complement rule:

$$P(X\ge 5)=1-P(X\le 4)=1-0.0006=0.9994$$

 

Command Ti83/84-calculator: $1-$binomcdf(12,0.8,4)

(b) Evaluate the definition of binomial probability at $k=7$:

 

 

P(X=7)​=C712​⋅0.77⋅(1−0.7)12−7=7!(12−7)!12!​⋅0.77⋅0.75≈0.0532

(c) Evaluate the definition of binomial probability at $k=0,1,2,3,4,5$:

 

$$\begin{array}{rl}P(X=0)& =C^0\cdot 0.7^0\cdot (1-0.7{)}^{12-0}=\genfrac{}{}{0ex}{}{12!}{0!(12-0)!}\cdot 0.7^0\cdot 0.7^{12}\approx 0.0000\\ P(X=1)& =C^1\cdot 0.7^1\cdot (1-0.7{)}^{12-1}=\genfrac{}{}{0ex}{}{12!}{1!(12-1)!}\cdot 0.7^1\cdot 0.7^{11}\approx 0.0000\\ P(X=2)& =C^2\cdot 0.7^2\cdot (1-0.7{)}^{12-2}=\genfrac{}{}{0ex}{}{12!}{2!(12-2)!}\cdot 0.7^2\cdot 0.7^{10}\approx 0.0000\\ P(X=3)& =C^3\cdot 0.7^3\cdot (1-0.7{)}^{12-3}=\genfrac{}{}{0ex}{}{12!}{3!(12-3)!}\cdot 0.7^3\cdot 0.7^9\approx 0.0001\\ P(X=4)& =C^4\cdot 0.7^4\cdot (1-0.7{)}^{12-4}=\genfrac{}{}{0ex}{}{12!}{4!(12-4)!}\cdot 0.7^4\cdot 0.7^8\approx 0.0005\\ P(X=5& =C^5\cdot 0.7^5\cdot (1-0.7{)}^{12-5}=\genfrac{}{}{0ex}{}{12!}{5!(12-5)!}\cdot 0.7^5\cdot 0.7^7\approx 0.0033\end{array}$$

 

Since it is not possible to obtain two different number of successes on the same simulation, we can use the addition rule for mutually exclusive events:

$$\begin{array}{rl}P(X<6)& =P(X\le 5)\\ & =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ & +P(X=4)+P(X=5)\\ & =0.0000+0.0000+0.0000+0.0001+0.0005+0.0033\\ & =0.0039\end{array}$$

 

 

 

Math output errorMath output error\begin{align*} P(X<6)&=P(X\leq 5) \\ &=P(X=0)+P(X=1)+P(X=2)+P(X=3) \\ &+P(X=4)+P(X=5) \\ &=0.0000+0.0000+0.0000+0.0001+0.0005+0.0033 \\ &=0.0039 \end{align*}

 

Math output errorMath output errorMath output error\begin{align*} P(X=0)&=C_0^{12} \cdot 0.7^0\cdot (1-0.7)^{12-0}=\dfrac{12!}{0!(12-0)!}\cdot 0.7^0\cdot 0.7^{12}\approx 0.0000 \\ P(X=1)&=C_1^{12} \cdot 0.7^1\cdot (1-0.7)^{12-1}=\dfrac{12!}{1!(12-1)!}\cdot 0.7^1\cdot 0.7^{11}\approx 0.0000 \\ P(X=2)&=C_2^{12} \cdot 0.7^2\cdot (1-0.7)^{12-2}=\dfrac{12!}{2!(12-2)!}\cdot 0.7^2\cdot 0.7^{10}\approx 0.0000 \\ P(X=3)&=C_3^{12} \cdot 0.7^3\cdot (1-0.7)^{12-3}=\dfrac{12!}{3!(12-3)!}\cdot 0.7^3\cdot 0.7^9\approx 0.0001 \\ P(X=4)&=C_4^{12} \cdot 0.7^4\cdot (1-0.7)^{12-4}=\dfrac{12!}{4!(12-4)!}\cdot 0.7^4\cdot 0.7^8\approx 0.0005 \\ P(X=5)&=C_5^{12} \cdot 0.7^5\cdot (1-0.7)^{12-5}=\dfrac{12!}{5!(12-5)!}\cdot 0.7^5\cdot 0.7^7\approx 0.0033 \end{align*}

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