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
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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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
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
Now,