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sp2606 [1]
2 years ago
12

An urn contains 3 red and 7 black balls. Players and withdraw balls from the urn consecutively until a red ball is selected. Fin

d the probability that selects the red ball. ( draws the first ball, then and so on. There is no replacement of the balls drawn.)
Mathematics
1 answer:
Sliva [168]2 years ago
7 0

Correct question:

An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn).

Answer:

The probability that A selects the red ball is 58.33 %

Step-by-step explanation:

A selects the red ball if the first red ball is drawn 1st, 3rd, 5th or 7th

1st selection: 9C2

3rd selection: 7C2

5th selection: 5C2

7th selection: 3C2

9C2 = (9!) / (7!2!) = 36

7C2 = (7!) / (5!2!) = 21

5C2 = (5!) / (3!2!) = 10

3C2 = (3!) / (2!) = 3

sum of all the possible events = 36 + 21 + 10 + 3 = 70

Total possible outcome of selecting the red ball = 10C3

10C3 = (10!) / (7!3!)

         = 120

The probability that A selects the red ball is sum of all the possible events divided by the total possible outcome.

P( A selects the red ball) = 70 / 120

                                         = 0.5833

                                         = 58.33 %

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Find the solution set of the following equations: <br><br>2y=2x+7<br><br>X=y+2
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Answer:

The system of equations has no solution.

Step-by-step explanation:

Given the system of equations

2y = 2x+7

x = y+2

solving the system of equations

\begin{bmatrix}2y=2x+7\\ x=y+2\end{bmatrix}

Arrange equation variables for elimination

\begin{bmatrix}2y-2x=7\\ -y+x=2\end{bmatrix}

Multiply -y+x=2 by 2:  -2y+2x=4

\begin{bmatrix}2y-2x=7\\ -2y+2x=4\end{bmatrix}

so adding

-2y+2x=4

+

\underline{2y-2x=7}

0=11

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\begin{bmatrix}2y-2x=7\\ 0=11\end{bmatrix}

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A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander aroun
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Answer:

The expected number of minutes the rat will be trapped in the maze is 21 minutes.

Step-by-step explanation:

The rat has two directions to leave the maze.

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If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability \frac{1}{3} after 2 minutes. And with probability \frac{2}{3} the rat will return to the starting point after 5 minutes of wandering.

Let <em>X</em> = number of minutes the rat will be trapped in the maze.

Compute the expected value of <em>X</em> as follows:

E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.

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2 years ago
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