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?
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Hi there!
Here's the answer:
•°•°•°•°•°<><><<><>><><>•°•°•°•°•°•
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
-----------------------------------------------
^^^^ 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
•°•°•°•°•°<><><<><>><><>•°•°•°•°•°•
Hope it helps
Here's the answer:
•°•°•°•°•°<><><<><>><><>•°•°•°•°•°•
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
-----------------------------------------------
^^^^ 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
•°•°•°•°•°<><><<><>><><>•°•°•°•°•°•
Hope it helps
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