GIVEN: A bag contain $5$ red marbles, $4$ green marbles and $1$ blue marble. A marble is chosen at random from the bag and not replaced, then a second marble is chose.

TO FIND: What is the probability both marbles are green.

SOLUTION:

Total marbles in bag $=10$

total number of green marbles $=4$

Probability that first marble will be green $P(A)=\frac{\text{total green marbles}}{\text{total marbles}}$

$=\frac{4}{10}=\frac{2}{5}$

Probability that second marble will be green $P(B)=\frac{\text{total green marble in bag}}{\text{total marble in bag}}$

$\frac{3}{9}=\frac{1}{3}$

As both events are disjoint

probability both marbles are green $=P(A)\times P(B)$

$=\frac{2}{5}\times\frac{1}{3}=\frac{2}{15}$

Hence the probability both marbles are green is $\frac{2}{15}$

8 0

3 years ago

Someone please help me by tonight I'm trying not to fail math. Also I will Brainliest if you help me get the right answer please

vredina [299]

Answer:

1. answer is D

2. Answer is D

3.Answer is B

4. Answer is C

5. Answer is B

6. Answer is D

7.Answer is D

Step-by-step explanation:

6 0

3 years ago