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A box contains 10 blue, 5 yellow and 8 orange balls. Two balls are drawn at random. What is the probability that none of the ball is yellow?

Sagot :

Syubbana2

A box contains 10 blue, 5 yellow and 8 orange balls. Two balls are drawn at random. The probability that none of the balls is yellow is [tex]\frac{153}{253}[/tex].

To solve this problem you can use the combination and probability formulas.

The Combination Formulas:

C(n, r) = n! ÷ (n - r)! r!

C(n, r) = [tex]\frac{n!}{(n-r)!r!}[/tex]

*noted:

n = the number of items

r = how many items are taken at a time

The Probability Formulas:

p(A) = n(A) ÷ n(S)

p(A) = [tex]\frac{n(A)}{n(S)}[/tex]

*Noted:

P(A) = the probability of an event “A”

n(A) = the amount taken

n(S) = the total number of events in the sample space.

Step-by-step explanation:

Given:

  • Amount of blue balls is 10 balls
  • Amount of yellow balls is 5 balls
  • Amount of orange balls is 8 balls
  • Two balls are drawn at random.

Question:

What is the probability that none of the balls is yellow?

Solution:

Step 1

Find the number of ways to get two blue balls.

C(n, r) = n! ÷ (n - r)! r!

C(10, 2) = 10! ÷ (10 - 2)! 2!

C(10, 2) = 10! ÷ 8! 2!

C(10, 2) = (10 x 9 x 8!) ÷ (8! x 2!)

C(10, 2) = (10 x 9) ÷ (2 x 1)

C(10, 2) = 90 ÷ 2

C(10, 2) = 45

Step 2

Find the number of ways to get two orange balls.

C(n, r) = n! ÷ (n - r)! r!

C(10, 2) = 8! ÷ (8 - 2)! 2!

C(10, 2) = 8! ÷ 6! 2!

C(10, 2) = (8 x 7 x 6!) ÷ (6! x 2 x 1)

C(10, 2) = (8 x 7) ÷ (2 x 1)

C(10, 2) = 56 ÷ 2

C(10, 2) = 28

Step 3

Find the number of ways to get one blue ball and one orange ball.

C(n, r) + C(n, r)

= C(10, 1) + C(8, 1)

= (10! ÷ (10 - 1)! 1!) x (8! ÷ (8 - 1)! 1!)

= (10! ÷ 9! 1!) x (8! ÷ 7! 1!)

= ((10 x 9!) ÷ (9! x 1)) x ((8 x 7!) ÷ (7! x 1))

= (10 ÷ 1) x (8 ÷ 1)

= 10 x 8

= 80

Step 4

Find the number of ways to get two the balls that none balls is yellow.

⇔ The number of two the balls or none balls is yellow is the number of two blue balls or two orange balls or one blue balls and one orange balls

⇒ The number of two the balls or none balls is yellow = 45 + 28 + 80

⇒ The number of two the balls or none balls is yellow = 153

n(A) = 153

Step 5

Find the number of ways to get two the balls.

C(n, r) = n! ÷ (n - r)! r!

C(23, 2) = 23! ÷ (23 - 2)! 2!

C(10, 2) = 23! ÷ 21! 2!

C(10, 2) = (23 x 22 x 21!) ÷ (21! x 2!)

C(10, 2) = (23 x 22) ÷ (2 x 1)

C(10, 2) = 506 ÷ 2

C(10, 2) = 253

n(S) = 253

Step 6

Find the probability that none of the balls is yellow.

p(A) = n(A) ÷ n(S)

p(A) = [tex]\frac{n(A)}{n(S)}[/tex]

p(A) = [tex]\frac{153}{253}[/tex]

So, the probability that none of the balls is yellow is [tex]\frac{153}{253}[/tex]

Learn more about:

  • A pot has 6 shiny balls and 14 matte balls. If a ball is removed from the pot without replacement: brainly.ph/question/29163450

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