A bucket cintains 8 red, 3 blue and 5 green marblea. if 4 marbles are drawn at random, what is the probabiloty that 2 are red and 2 are blue?
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Total number of marbles = 16
P ( red) = 8/16 => P = 2
P (blue) = 3/16
Hope it helps!!
Cheers!!
P ( red) = 8/16 => P = 2
P (blue) = 3/16
Hope it helps!!
Cheers!!
Answered by
3
Hey........!!! here is ur answer......☺️☺️☺️
Given that....there r 8 red, 3 blue, and 5 green marbles in the bucket.
Then total marbles = 8+3+5 = 16
If 4 marbles are drawn out randomly.
then probability of 2 marbles are red and 2 marbles are blue = (8C2×3C2)/16C2
=> [8!/2!(8–2)! × 3!/2!(3–2)!]/16!/2!(16–2)!
=>{(8!/2!6!)×(3!/2!1!)}/16!/2!14!
=>{(8×7)/2×(3/1)}/16×15/2
Because we know that n! = n(n–1)(n–2)......
=>[(4×7)×3]/(8×15)
=>21/30
=>7/10
I hope it will help you........✌️✌️✌️
Given that....there r 8 red, 3 blue, and 5 green marbles in the bucket.
Then total marbles = 8+3+5 = 16
If 4 marbles are drawn out randomly.
then probability of 2 marbles are red and 2 marbles are blue = (8C2×3C2)/16C2
=> [8!/2!(8–2)! × 3!/2!(3–2)!]/16!/2!(16–2)!
=>{(8!/2!6!)×(3!/2!1!)}/16!/2!14!
=>{(8×7)/2×(3/1)}/16×15/2
Because we know that n! = n(n–1)(n–2)......
=>[(4×7)×3]/(8×15)
=>21/30
=>7/10
I hope it will help you........✌️✌️✌️
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