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timurjin [86]
4 years ago
7

A square is inscribed in the circle. A point in the figure is selected at random. Find the probability that the point will be in

the part that is NOT shaded.
Mathematics
1 answer:
Snezhnost [94]4 years ago
4 0
Ok 
Probability (the point will be on unshaded part)  = area of unshaded part / area of the circle.

=   pi r^2 - s^2
     --------------            where r = radius of circle and s = length of the side of the square.
        pi r^2
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see below

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A bag contains five white balls and five black balls. Your goal is to draw two black balls.
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a.) the probability that both the ball drawn are black  = \frac{10}{45} = \frac{2}{9}

b.) Sum of all the probabilities = \frac{\frac{2}{9} }{1 - \frac{2}{9} }  = \frac{\frac{2}{9} }{\frac{8}{9} }  = \frac{2}{8} =\frac{1}{4} = 0.25

Step-by-step explanation:

a) two balls are drawn at random .

   total number of balls   =  10

 number of black balls    =  5

  number of white balls    = 5.

   the number of ways two balls can be drawn = \binom{10}{2}  = \frac{10!}{2! 8!}  = \frac{10\times9}{2}  = 45

  the number of ways two black balls can be drawn = \binom{5}{2}  = \frac{5!}{2! 3!}  = \frac{5\times4}{2}  = 10

   the probability that both the ball drawn are black  = \frac{10}{45} = \frac{2}{9}

b) probability that First draw of two black balls = \frac{2}{9}

  probability that first draw is of two white balls is and second draw is of

  two black balls  = \frac{2}{9} \times\frac{2}{9}

  Probability that first two draws are of white balls and the third draw is of

  two black balls = \frac{2}{9} \times\frac{2}{9} \times\frac{2}{9}

  This is a geometric sequence and the final probability will be the sum of    

  all such probabilities, where we can take the the sequence to be an infinite series and the first term is \frac{2}{9} and the common ratio is \frac{2}{9}  which is less than 1.

  Sum of all the probabilities = \frac{\frac{2}{9} }{1 - \frac{2}{9} }  = \frac{\frac{2}{9} }{\frac{8}{9} }  = \frac{2}{8} =\frac{1}{4} = 0.25

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4 years ago
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