There are 4 white tokens and 6 blue tokens in a bag. Once a token is selected, it is not replaced. Find the probability of selecting two white tokens

The probability of selecting two white tokens without replacement is .

Given:

The number of white tokens is 4.

The number of blue tokens is 6.

To find:

The probability of selecting two white tokens without replacement.

Explanation:

We know that,

The probability of getting first white token is:

Now, the number of remaining tokens is 9 and the number of remaining white tokens is 3. So, the probability of getting second white token is:

The probability of selecting two white tokens without replacement is:

Therefore, the probability of selecting two white tokens without replacement is .

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2/15

Step-by-step explanation: To find this compound probability, first note that there is no replacement.

The probability of drawing the first white token is 4/10 or 2/5.

The probability of drawing the second token is 3/9 or 1/3 because one white token has already been taken away. Now, multiply 2/5 by 1/3 to get the answer, which is 2/15

3x² * (6x² + 4x - 8) = 18x⁴ + 12x³ - 24x²

2x * (9x³ + 6x² - 12x) = 18x⁴ + 12x³ - 24x²

x² * (18x² + 12x - 24) = 18x⁴ + 12x³ - 24x²

6x² * (3x² + 2x - 4) = 18x⁴ + 12x³ - 24x²

Step-by-step explanatation:

Let's set unknown variables as y (and y₁, y₂, etc)

Question 1:

y *  (6x² + 4x - 8) = 18x⁴ + 12x³ - 24x²

So, divide both sides by (6x² + 4x - 8)

y = 3x²

Question 2:

2x * (9x^y₁ + y₂*x^y₃ - y₄*x) = 18x⁴ + 12x³ - 24x² '

Divide both sides by 2x.

(9x^y₁ + y₂*x^y₃ - y₄*x) = 9x³ + 6x² -12x

y₁ = 3, y₂ = 6, y₃ = 2, y₄ = 12

Question 3:

x^y * (18x² + 12x - 24) = 18x⁴ + 12x³ - 24x²

Divide both sides by (18x² + 12x - 24)

x^y = x²

Question 4:

y₁  * x^y₂ * (3x² + 2x - 4) = 18x⁴ + 12x³ - 24x²

Divide both sides by (3x² + 2x - 4)

y₁  * x^y₂ = 6x²

y₁ = 6, y₂ = 2

I hope this helps! Feel free to ask any questions!