Question

a bag contains 3 red 4 blue 5 green balls if three balls are picked up at random what is the probability that all are blue​

Answer 1

he possibilities are ,

• A {1st draw red, 2nd draw green},
• B{1st draw blue, 2nd draw green} &
• C{1st draw green, 2nd draw is also green}.

Now,

• P(A) = 3/12 * 5/12 = 15/144
• P(B)= 4/12 * 5/12 = 20/144
• P(C)= 5/12 * 5/12 = 25/144.

Hence, required probability is P(A)+P(B)+P(C)= 60/144=15/36

Answer 2

Total number of balls = 12 [3+4+5]

Number of blue balls = 4

we know that,

$probability = \frac{no.ofballs}{total \: no.ofballs}$

So, probability is

$\frac{4}{12} = \frac{1}{3}$

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