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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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As all of the bundles have the same content inside, so assuming that there is x number of Fiction books and y number of Non-fiction books in each bundle.

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There are 64 Fiction books, so

nx=64 ...(i)

Or x=64/n ...(ii)

also, there are 48 Non-Fiction books, so

ny=48 ...(iii)

Or y=48/n ...(iv)

Observing that the numbers x, y, and n are counting numbers and from equations (i) and (iii), n is the common factor of 64 and 48.

The possible common factors of 64 and 48 are,

n=1,2,4,8, and 16.

So, my team can send 1,2,3,4,8 or 16 bundles of books.

Now, from equations (ii) and (iv),

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For n=2:

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y=48/2=48

So, for 2 bundles, the number of Fiction and Non-fictions books are 32 and 24 respectively.

For n=4:

x=64/4=16

y=48/4=12

So, for 4 bundles, the number of Fiction and Non-fictions books are 16 and 12 respectively.

For n=8:

x=64/8=8

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So, for 8 bundles, the number of Fiction and Non-fictions books are 8 and 6 respectively.

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Let's substitute 9 - d for q in the equation immediately above:

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