Question

from a bag containing 10 red, 4blue and 6 black balls, a ball is drawn at random. find the probability of drawing a)a red ball b)a blue or black ball c)not a black ball​

Answer 1

Answer:

total no of events = 10+4+6 = 20

a) p(e) = 10/20= 1/2

b) p(e) = 4+10 =14

14/20 = 7/10

C) p(e) = 20-6 = 14

14/20 = 7/10

Answer 2

a) $\dfrac{1}{2}$ b) $\dfrac{1}{2}$, c) $\dfrac{7}{10}$

Step-by-step explanation:

Since we have given that

Number of red balls = 10

Number of blue balls = 4

Number of black balls = 6

So, a) Probability of drawing a red ball is given by

$\dfrac{10}{20}=\dfrac{1}{2}$

b) probability of drawing a blue or black ball is given by

$\dfrac{4}{20}+\dfrac{6}{20}=\dfrac{10}{20}=\dfrac{1}{2}$

c) Probability of not getting a black ball is given by

$1-P(black)\\\\=1-\dfrac{6}{20}\\\\=1-\dfrac{3}{10}\\\\=\dfrac{10-3}{10}\\\\=\dfrac{7}{10}$

Hence, a) $\dfrac{1}{2}$ b) $\dfrac{1}{2}$, c) $\dfrac{7}{10}$

# learn more:

A urn contains 6 red 4blue 2green 3yellow marbles if three marbles are picked up at random what is probabilty that 2 are blue and one is yellow

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