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BabaBlast [244]
3 years ago
5

A certain university has 10 vehicles available for use by faculty and staff. four of these are vans and 6 are cars. on a particu

lar day, only two requests for vehicles have been made. suppose that the two vehicles to be assigned are chosen in a completely random fashion from among the 10. (a) let e denote the event that the first vehicle assigned is a van. what is p(e)? 0.4 correct: your answer is correct. (b) let f denote the probability that the second vehicle assigned is a van. what is p(f | e)? incorrect: your answer is incorrect. (c) use the results of parts (a) and (b) to calculate p(e and f) (hint: use the definition of p(f | e).)
Mathematics
1 answer:
REY [17]3 years ago
6 0
<span>There are 4 vans. So we have that probability that the first vehicle is a van p (e) = 4/10 = 0.4. P(e|f) = P( f and e) / p (f) p(e) = 0.4 and p(f) = 3/9 P (f and e) = 0.40 * 0.33 = 0.132 So p(e|f) = 0.4 * 0.33/ 0.33 = 0.4 P(f and e) p(f) * p(e) = 0.4 * 0.33 = 0.132</span>
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Then, solving the determinant

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In our case,

\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}

Where we used the formula for AxB to calculate ixj.

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\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}

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