Question

Plssss help

A bag contains 6 red marbles, 10 white marbles, and 6 blue marbles. You draw 3 marbles out at random, without replacement.

A) What is the probability that all the marbles are red?

B) what is the probability the exactly 2 of marbles red?

C) What is the probability that none of the marbles are red? ​

Answer 1

A) 1/77; B) 12/77; C) 4/11

Explanation:

A) There are a total of 22 marbles. 6 of them are red.

On the first draw, the probability of getting a red marble is 6/22.

On the second draw, there's one less red marble and one less marble total, so the probability of getting another red marble is 5/21.

Similarly, on the third draw, the probability of getting a red marble is 4/20.

So the probability that all three draws are red marbles is:

P = (6/22) (5/21) (4/20)

P = 1/77

Another way this can be calculated is with combinations:

P = (ways to choose 3 red marbles from 6) / (ways to choose 3 marbles from 22)

P = ₆C₃ / ₂₂C₃

P = 20 / 1540

P = 1/77

B) The same logic can be repeated here. Using the first method, if the first two selection are red:

P = (6/22) (5/21) (16/20) = 4/77

If the first and third are red:

P = (6/22) (16/21) (5/20) = 4/77

If the last two are red:

P = (16/22) (6/21) (5/20) = 4/77

So the total probability is:

P = 4/77 + 4/77 + 4/77

P = 12/77

Using the second method:

P = (ways to choose 2 red from 6) × (ways to choose 1 non-red from 16) / (ways to choose 3 from 22)

P = ₆C₂ ₁₆C₁ / ₂₂C₃

P = 15 × 16 / 1540

P = 12/77

C)

Same logic:

P = (16/22) (15/21) (14/20)

P = 4/11

Or:

P = ₁₆C₃ / ₂₂C₃

P = 560 / 1540

P = 4/11

Answer 2

• A)
  `\displaystyle (1)/(77)`.
• B)
  `\displaystyle (12)/(77)`.
• C)
  `\displaystyle (4)/(11)`.

Explanation:

All marbles here are identical. Also, the question isn't concerned about the order in which the marbles are drawn. Thus, all calculations here shall be combinations rather than permutations.

A)

How many ways to choose three out of six identical red marbles without replacement?

`\displaystyle _6C_3 = c(6, 3) = {6\choose 3} = 20`.

Note that these three expressions are equivalent. They all represent the number of ways to choose 3 out of 6 identical items without replacement.

How many ways to choose three out of all the 6 + 10 + 6 = 22 marbles?

`\displaystyle _(22) C_(3) = 1540`.

The probability of choosing three red marbles out of these 22 marbles will be:

`\displaystyle \frac{\text{Number of ways for choosing three out of six red marbles}}{\text{Number of ways to choose three out of 22 marbles}} = (20)/(1540) = (1)/(77)`.

B)

How many ways to choose two out of six identical red marbles without replacement?

`\displaystyle _6 C_2 = 15`.

How many ways to choose one out of 10 + 6 = 16 non-red marbles?

`_(16) C_1=16`.

Choosing two red marbles does not influence the number of ways of choosing a non-red marble. Both event happen and are independent of each other. Apply the product rule to find the number of ways of choosing two red marbles and one non-red marble out of the pile of 22.

`_6 C_2 \cdot _(16) C_1= 240`.

Probability:

`\displaystyle (240)/(1540) = (12)/(77)`.

Double check that the order doesn't matter here.

C)

None of the marbles are red. In other words, all three marbles are chosen out of a pile of 10 + 6 = 16 white and blue marbles. Number of ways to do so:

`_(16) C_(3) = 560`.

Probability:

`\displaystyle (560)/(1540)= (4)/(11)`.