When choose 3 out of 8 books, the probability of the most ex

Question

Please help me to understand this... I am always terrified when i face PROBABILITY questions...

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Please helpe me kindly and thank you in advance...

pretttyune · Answer

yes... the question is .. In choosing 3 out of 8 books,  the probability  of the most expensive two books must be included?

marcodonzelli · Answer

it is: consider the most expensive 2 books....than we have 6 slots free to combine the books. the total number of combs is 8C3 because we have a total of 8 books to combine in groups of three...prob=6/8*7=3/28

pmenon · Answer

i thought about it like this.

probability = desired outcomes / total out comes

where desired outcomes is the number of outcomes of picking 3 out of 8 books, but the two most expensive have to be in there, so ive got (1C1)*(1C1)*(6C1) / 8C3 = 6/56 = 3/28

bmwhype2 · Answer

i did it slightly differently.

(2/8 *1/7 * 6/6) * 3c1 = 3/28

AlexBon · Answer

Walker pls elucidate why not 2!6c1/8c3

walker · Answer

p=\frac{C^2_2*C^6_1}{C^8_3}=\frac{1*6*3*2}{8*7*6}=\frac{3}{28}

or

p=\frac{C^2_2*C^6_1*P^3_3}{P^8_3}=\frac{1*6*3*2}{8*7*6}=\frac{3}{28}

or

p=\frac28*(\frac17*\frac66+\frac67*\frac16)+\frac68*\frac27*\frac16=\frac{3*12}{8*7*6}=\frac{3}{28}

I think 2! is permutation:  P^2_2 . But you use  C^8_3  for all variants that mean  AB C and  BA C are the same variant. If we distinguish between ABC and BAC, we use  P^8_3  and  P^3_3

Hope this help.

srivas · Answer

Q. When choose 3 out of 8 books, the probability of the most expensive two books must be included?

Soln: 6C1/8C3

jeeteshsingh · Answer

Out of 8 .. 2 books (which are most exp) should be considered in your selection.

No of ways the 2 Exp books can be selected = 2c2 = 1

remaining we need to choose one more book from the 6 left over books = 6c1 = 6

Total ways of selecting 3 books out of 8 = 8c3

Prob = 1X6 / 8c3 = 3/28

ButwhY · Answer

Probability of selecting most expensive out of 8 =1/8
Probability of selecting next most expensive out of 7 =1/7
And out of 6 left out can select any one out of 6 = 6c1/6c1
now probability of selecting 2 most expensive = 1/8*1/7*1*3!
=3/8(3! is because of arrangements)

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When choose 3 out of 8 books, the probability of the most ex

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