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There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them are purple, and 24 of them are blue. Suppose Hao draws 22 balls from the box with replacement (he draws the ball, records its color, and then puts it back into the box). Find the probability that he draws 2 red balls, 5 green balls, 10 purple balls, and 5 blue balls

Respuesta :

Answer:

1.1348621526 X 10⁻¹³

Step-by-step explanation:

Number of Red balls=25

Number of Green balls=19

Number of Purple balls=30

Number of Blue=24

Total=25+19+30+24=98

Since the balls are picked with replacements, probability of picking a color will be the same all true.

P(Picking a red ball)= 25/98

P(Picking a green ball)= 19/98

P(Picking a purple ball)= 30/98

P(Picking a blue ball)= 24/98

P(2 red balls)= 25/98 X 25/98 = (25/98)²

P(5 green balls)= 19/98 X 19/98 X 19/98 X 19/98 X 19/98 =(19/98)⁵

P(10 purple balls)= 30/98 X 30/98 X 30/98 X 30/98 X 30/98 X 30/98 X 30/98 X 30/98 X 30/98 X 30/98 =(30/98)¹⁰

P( 5 blue balls) =24/98 X 24/98 X 24/98 X 24/98 X 24/98 =(24/98)⁵

P(2 red balls, 5 green balls, 10 purple balls, and 5 blue balls) = (25/98)² X (19/98)⁵ X (30/98)¹⁰ X (24/98)⁵

=1.1348621526 X 10⁻¹³

Answer:

Probability = 1.0015 * 10 ^ -3

Step-by-step explanation:

Number of Red balls=25

Number of Green balls=19

Number of Purple balls=30

Number of Blue=24

Total=25+19+30+24=98

He draws 22 balls with replacement.

So arrangement doesn't matter.

Probability of 2 red balls, 5 green balls, 10 purple balls, and 5 blue balls is =

( 25C2 * 19C5 * 30C10 *24C5)/98C22

Note the C represent combination.

=( 300*11628*30045015*42504)/(4.448120577*10^21)

= 1.0015 * 10 ^ -3

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