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Papessa [141]
3 years ago
6

Suppose that you are seated next to a stranger on an airplane and you start discussing various topics such as where you were bor

n (what state or country), what your favorite movie of all time is, your spouse's occupation, and so on. For simplicity, assume that the probability that your details match for any given topic is 1 50 and is independent from one topic to the next. If you discuss 17 topics, how surprising would it be to find that you match on at least one of them? (Round your answer to four decimal places.)
Mathematics
1 answer:
Marat540 [252]3 years ago
4 0

Answer:

0.2907

Step-by-step explanation:

Given that you are seated next to a stranger on an airplane and you start discussing various topics such as where you were born (what state or country), what your favorite movie of all time is, your spouse's occupation, and so on.

Since each topic is independent we find that probability for success in each topic = constant = 1/50 = 0.02

X no of topics that match is binomial with n = 17 and p = 0.02

Required probability

= Probability to find that you match on at least one of them

=P(X\geq 1)\\=1-P(X=0)\\=1-(1-0.02)^{17} \\=0.2907

It would not be surprising as probability is reasonably large.

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<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}

Let us bring the equations in same form for sake of simplicity in comparison

\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}

Now we have two equations  

\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}

Let us first see what is requirement for system of equations have an infinite number of solutions

If  a_{1} x+b_{1} y+c_{1}=0 and a_{2} x+b_{2} y+c_{2}=0 are two equation  

\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} then the given system of equation has no infinitely many solutions.

In our case,

\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}

 As for infinitely many solutions \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}

Hence b must be equal to -6 for infinitely many solutions for system of equations y = 6x + b and  -3 x+\frac{1}{2} y=-3

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