Math, asked by Ramitsaini911, 10 months ago

A BOX CONTAIN 4 RED BALL AND 12 BLACK BALLS . FIND THE PROBABILITY THAT THE BALL (a) a red ball and (b)a black ball chosen at random from the box ,also prove that sum of these two probability is one.​

Answers

Answered by kingabuzar77
1

Step-by-step explanation:

No. of Red Balls = 4 balls

No. of Black balls = 12 balls

Total no. of balls = 16 balls

a) P(Red ball ) = No. of Red balls ÷ Total no.of balls

= 4÷16

= 0.25

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b) P(Black ball) = No. of black ball÷Total no. of balls

= 12÷16

=0.7

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Answered by siddhivd16
14

Answer:

The box contains 4 red balls, 12 black balls.

S = { R1, R2, R3, R4, B1, B2, B3, B4, B5, B6, B7, B8, B9, B10, B11, B12}

n(S) = 16

Step-by-step explanation:

A) Let A be the event that the ball is red.

A = { R1, R2, R3, R4}

n(A) = 4

p(a) =  \frac{n(a)}{n(s)}  \\  \ = \:  \frac{4}{16}  \\  =  \frac{1}{4}

B) Let B be the event that ball is black.

B = { B1, B2, B3, B4, B5, B6, B7, B8, B9, B10, B11, B12}

n(B) = 12

p(b) =  \frac{n(b)}{n(s)}  \\  =  \frac{12}{16}  \\  =  \frac{3}{4}

Now,

p(a) + p(b)  \\  =  \frac{1}{4}  +  \frac{3}{4}  \\  =  \frac{1 + 3}{4 }  \\  =  \frac{4}{4}  \\  = 1

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