Question

What is the probability of drawing two yellow marbles if the first one is NOT placed back into the bag before the second draw? Their is 10 marbles total, 2 yellow, 3 pink, and 5 blue

Answer 1

Break it down into two parts. First, what is the probability of drawing a blue marble on the first draw?Since there are 5 blue marbles and 10 total, the probability is 5⁄10, or 1/2. Now since we no longer have that blue marble, there are 4 blue marbles and 9 total. The chances of drawing a blue marble are 4/9. Therefore, the chance that both marbles drawn are blue is the chance that the first one is blue times the chance that the second one is blue. 1/2 * 4/9 = 4/18 = 2/9 Remember, math is always trying to trick you. It wants you to try and do the whole big problem at once, which can be difficult. Break it down into smaller problems, then use your answers to small parts to find the answer to the big question.

Ishrat

Answer 2

Answer:

$\frac{4}{45}$

Step-by-step explanation:

We are given that

Total number of marbles=10

Number of yellow marble=2

Number of pink marble=3

Number of blue marble=5

We have to find the probability of drawing two yellow  marbles if the first one is not placed into the bag before the second draw.

Probability: P(E)=$\frac{Number\;of\;favorable\;cases}{total\;number\;of\;cases}$

The probability of drawing first yellow marble=$\frac{2}{10}$

The probability of drawing first yellow marble=$\frac{1}{5}$

When first yellow marble is not placed back into the bag before the second draw.

Then,number of remaining  yellow marble=1

Total number of marbles=9

The probability of drawing second yellow marble =$\frac{4}{9}$

Now, the probability of drawing two yellow marbles =$\frac{1}{5}\times \frac{4}{9}=\frac{4}{45}$

Hence, the probability of drawing two yellow marbles=$\frac{4}{45}$