Question

A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen?

Answer 1

Hi there!
Here's the answer:

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In a Bag

No. of Blue Balls = 4
No. of Red Balls = 2
No. of Black Balls = 3
Total No. of Balls = 9

Let A be the event that a Blue pen is drawn
& B be the Event that a Black pen is drawn.

Out of 3 Trials, Occurrence of 2 A's and 1 B is to be found.

This can happen in following ways:

AAB, ABA, BAA.

•°• The Required Probability =3 ×(4/9)×(4/9)×(3/9)

= (3×16×3)/(9×9×9)

= 16/81

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^^^^ Here,
P(A) = P(drawing a Blue pen) = 4/9
P(B) = P(drawing a Black pen) = 3/9.

Since the successive trials are conducted with Replacement, P(A) & P(B) remain unaffected in successive trials

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Hope it helps