• February 17th 2010, 11:53 AM
john002

For [ 2x^2 - 1/x]^6, determine
a) The middle term
b) The term containing x^3

Given the universal set S = {-3,-2,-1,0,1,2,3,4} and the sets A = {1,2,3,4}, B = {-3,-2,-1}, find C = {-2,-1,0,1,2}
FInd
a) A'
b) (A n C) and (B n C)
c) (B n C)'

Your are dealt one card from a standard deck of playing cards.
a) What is the probability that the card is 5?
b) What is the probability that the card is a 8 or a club?

Eight percent of the objects produced by a machine are defective. If you examine 50 objects produced by the machine, find the probability that
a) 3 are defective
b) More than 3 are defective

A class consists of 8 boys and 5 girls
a) In how many ways can a teacher choose a committee of 3 from this class?
b) How many of these committees will contain at least 1 boy?
c) How many will contain exactly 1 girl
What is the probability that number chosen from 1 to 500 is divisible by 2, 5 or 6?

A box contains 5 black balls, 4 red balls, 4 green balls, and 5 blue balls.
a) YOu select 6 balls from the box without replacing them. What is the probabilty that
you will select excatly 3 black balls?
b) You select 6 balls from the box without replacing them. What is the probability that
you draw at least 1 black balls?
• February 17th 2010, 10:29 PM
Soroban
Hello, john002!

Quote:

For $\left(2x^2 - \tfrac{1}{x}\right)^6$, determine

(a) the middle term

(b) the term containing $x^3$

We are expected to know the Binomial Theorem.

$\left(2x^2 - \tfrac{1}{x}\right)^6 \;=\;{6\choose6}(2x^2)^6\left(\text{-}\tfrac{1}{x}\right)^0 + {6\choose5}(2x^2)^5\left(\text{-}\tfrac{1}{x}\right)^1 + {6\choose4}(2x^2)^4\left(\text{-}\tfrac{1}{x}\right)^2 +$ ${6\choose3}(2x^2)^3\left(\text{-}\tfrac{1}{x}\right)^3 + {6\choose2}(2x^2)^2\left(\text{-}\tfrac{1}{x}\right)^4 + {6\choose1}(2x^2)^1\left(\text{-}\tfrac{1}{x}\right)^5 + {6\choose0}(2x^2)^0\left(\text{-}\tfrac{1}{x}\right)^6$

. . . . . . . $=\; (1)(64x^{12})(1) \;+\; 6(32x^{10})\left(\text{-}\tfrac{1}{x}\right)\;+\;15(16x^8)\left(\tfrac{1}{ x^2}\right) \;+ \;20(8x^6)\left(\text{-}\tfrac{1}{x^3}\right) \;+\; 15(4x^4)\left(\tfrac{1}{x^4}\right)$ . $+\;6(2x^2)\left(\text{-}\tfrac{1}{x^5}\right) + 1(1)\left(\tfrac{1}{x^6}\right)$

. . . . . . . $=\;64x^{12} - 192x^9 + 240x^6 \underbrace{-\; 160x^3}_{(a),\;(b)} + 60 - \frac{12}{x^3} + \frac{1}{x^6}$

Quote:

You are dealt one card from a standard deck of playing cards.

a) What is the probability that the card is 5?

b) What is the probability that the card is a 8 or a club?

There are 52 cards in a standard deck of cards.

(a) Four of them are 5's.

. . $P(5) \:=\:\frac{4}{52} \:=\:\frac{1}{13}$

(b) There are four 8's: . $8\heartsuit,\;8\spadesuit,\;8\diamondsuit,\;8\club suit$
. . .There are 13 Clubs: . $A\clubsuit,\;2\clubsuit,\;3\clubsuit,\;4\clubsuit, \;5\clubsuit,\;6\clubsuit,\;7\clubsuit,\;8\clubsui t,\;9\clubsuit,\;10\clubsuit,\;J\clubsuit,\;Q\club suit.\;K\clubsuit$

. . .But don't count the $8\clubsuit$ twice!

Hence, there are 16 cards that are an 8 or a Club.

. . $P(\text{8 or Club}) \:=\:\frac{16}{52} \:=\:\frac{4}{13}$

Quote:

A box contains 5 black balls, 4 red balls, 4 green balls, and 5 blue balls.
You select 6 balls from the box without replacing them.

(a) What is the probabilty that you select excatly 3 black balls?

There are: . ${18\choose6} \,=\,18,\!564$ possible outcomes.

We have: .5 Black balls and 13 Others
We want: 3 Black balls and 3 Others.

There are: . ${5\choose3} = 20$ ways to pick 3 Black.

There are: . ${13\choose5} = 286$ ways to pick 3 Others.

Hence, there are: . $20 \times 286 \:=\:5,\!720$ ways to pick 3 Blacks and 3 Others.

Therefore: . $P(\text{at least 3 Blacks}) \:=\:\frac{5,\!720}{18,\!564} \;=\;\frac{110}{357}$

Quote:

(b) What is the probability that you draw at least 1 black ball?
The opposite of "at least one Black" is "no Blacks".

To get no Blacks, we want 6 of the 13 Others: . ${13\choose6} = 1716$ ways.
. . Hence: . $P(\text{no Blacks}) \:=\:\frac{1716}{18,\!564} \:=\:\frac{11}{119}$

Therefore: . $P(\text{at least one Black}) \;=\;1 - \frac{11}{119} \;=\;\frac{108}{119}$