Mathematics, 20.02.2020 22:38 Weser17

Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size four is equally likely, find the probabilities of the following events:1. The committee consists of two men and two women.2. The committee has more women than men.3. The committee has at least one man. For the remainder of the problem, assume that Alice and Bob are among the ten people being considered.4. Both Alice and Bob are members of the committee. Shouldn't it be like a coin toss but with more than just heads and tails?

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Answer from: marshalldayton1452

1. 10/21

2. 11/42

3. 41/42

4. 2/15

Step-by-step explanation:

Probability defines the ratio of the number of favorable outcomes to the total number of outcomes.

Combination Operations

$5C2 = \frac{5! }{2! (5-2)!} = \frac{5*4*3! }{2! 3!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10$

$10C4 = \frac{10! }{4! (10-4)!} = \frac{10*9*8*7*6! }{4! 6!} =\frac{10*9*8*7 }{4*3*2*1} =\frac{5040}{24}= 210$

$5C3 = \frac{5! }{3! (5-3)!} = \frac{5*4*3! }{3! 2!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10$

$5C1 = \frac{5! }{1! (5-1)!} = \frac{5*4! }{1! 4!} =\frac{5 }{1} = 5$

$8C2 = \frac{8! }{2! (8-2)!} = \frac{8*7*6! }{2! 6!} =\frac{8*7 }{2*1} = 4*7= 28$

$5C4 = \frac{5! }{4! (5-4)!} = \frac{5*4! }{4! 1!} =\frac{5 }{1} = 5$

$5C0 = \frac{5! }{0! (5-0)!} = \frac{5! }{0! 5!} = 1$

1. Probability of forming a committee consisting of two men and two women

= $\frac{5C2 * 5C2}{10C4} =\frac{10*10}{210} =\frac{10}{21}$

2. Probability of forming a committee that has more women than men

(NOTE: Since the committee desired has only 4 members, forming a committee that has more women than men can only be possible if we have 4 women, 0 men or 3 women 1 man)

= $\frac{5C3 * 5C1 + 5C4*5C0}{10C4} = \frac{10 * 5 + 5*1 }{210} = \frac{50 + 5}{210}=\frac{55}{210} = \frac{11}{42}$

3. Probability of forming a committee that has at least one man

(NOTE: at least one man implies the committee must not have all women members )

= $1 - \frac{5C4}{10C4} = 1 - \frac{5}{210} = 1 - \frac{1}{42} = \frac{41}{42}$

4. Probability of forming a committee with both Alice and Bob are members of the committee (NOTE: If Alice and Bob are compulsory member of the committee, there are only two more people to join the committee from the remaining 8 people)

$=\frac{8C2}{10C4} =\frac{28}{210} =\frac{2}{15}$

it be like a coin toss ut a little more complicated. For every toss we have 4 outcomes from 210 possililities

Answer from: Quest

$y> \frac{1}{3}x$ and y≤ -x +4 .

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be = $\frac{1-0}{3-0} = \frac{1}{3}$ .

y-intercept of dotted line is y=0 only.

plugging slope m= $\frac{1}{3}$ and b=0 in slope-intercept form y=mx+b, we get

$y=\frac{1}{3}x$ .

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

$y> \frac{1}{3}x$ .

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be = $\frac{4-1}{0-3} = \frac{3}{-3}=-1$ .

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be

y≤ -x +4 .
Write a system of inequalities to represent the shaded portion of the graph.

Answer from: Quest

Ithink the right answer could be 50%

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Out of five men and five women, we form a committee consisting of four different people. Assuming th...