Question

a box contains 9 red balls , 5blue balls, 6 green balls. three balls are drawn out, what is the probability that at least one ball is green​

Answer 1

The probability that at least one ball is green is 194/285

Given :

• A box contains 9 red balls , 5 blue balls, 6 green balls.

• Three balls are drawn out

To find :

The probability that at least one ball is green

Solution :

Step 1 of 3 :

Find total number of possible outcomes

Here it is given that the box contains 9 red balls , 5 blue balls, 6 green balls.

Now three balls are drawn out

So total number possible outcomes

$\displaystyle \sf{ = {}^{20}C_3 }$

$\displaystyle \sf{ = \frac{20!}{3! \: (20 - 3)!} }$

$\displaystyle \sf{ = \frac{20!}{3! \: 17!} }$

$\displaystyle \sf{ = \frac{20 \times 19 \times 18 \times 17!}{3! \: 17!} }$

$\displaystyle \sf{ = \frac{20 \times 19 \times 18 }{3 \times 2} }$

$\displaystyle \sf{ =1140 }$

Step 2 of 3 :

Find the probability that no ball is green

Let A be the event that no ball is green

Total number of balls which are not green

= 9 + 5

= 14

Then total number of possible outcomes for the event A

$\displaystyle \sf{ = {}^{14}C_3 }$

$\displaystyle \sf{ = \frac{14!}{3! \: (14 - 3)!} }$

$\displaystyle \sf{ = \frac{14!}{3! \: 11!} }$

$\displaystyle \sf{ = \frac{14 \times 13 \times 12 \times 11!}{3! \: 11!} }$

$\displaystyle \sf{ = \frac{14 \times 13 \times 12 }{3! } }$

$\displaystyle \sf{ = \frac{14 \times 13 \times 12 }{3 \times 2 } }$

$\displaystyle \sf{ = 364 }$

So the probability that no ball is green

= P(A)

$\displaystyle \sf{ = \frac{364}{1140} }$

Step 3 of 3 :

Find the probability that at least one ball is green

Hence the required probability that at least one ball is green

= 1 - P(A)

$\displaystyle \sf{ = 1 - \frac{364}{1140} }$

$\displaystyle \sf{ = 1 - \frac{91}{285} }$

$\displaystyle \sf{ = \frac{285 - 91}{285} }$

$\displaystyle \sf{ = \frac{194}{285} }$

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