A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a black and then a red marble out of the bag without replacing the black marble before drawing the red marble? Answer options with 5 options A. StartFraction 1 over 144 EndFraction B. StartFraction 1 over 11 EndFraction C. StartFraction 1 over 12 EndFraction D. StartFraction 7 over 12 EndFraction E. StartFraction 27 over 44 EndFraction
Question
A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a black and then a red marble out of the bag without replacing the black marble before drawing the red marble?
Answer options with 5 options A. StartFraction 1 over 144 EndFraction
B. StartFraction 1 over 11 EndFraction
C. StartFraction 1 over 12 EndFraction
D. StartFraction 7 over 12 EndFraction
E. StartFraction 27 over 44 EndFraction
Solution
To solve this problem, we need to calculate the probability of both events happening (drawing a black marble and then a red marble) and then multiply those probabilities together.
Step 1: Calculate the total number of marbles. There are 3 black, 4 red, 3 yellow, and 2 green marbles. So, the total number of marbles is 3 + 4 + 3 + 2 = 12 marbles.
Step 2: Calculate the probability of drawing a black marble. The probability of drawing a black marble is the number of black marbles divided by the total number of marbles. So, the probability is 3/12 = 1/4.
Step 3: Calculate the probability of drawing a red marble after drawing a black marble. After drawing a black marble, there are now 11 marbles left in the bag. The probability of drawing a red marble is the number of red marbles divided by the new total number of marbles. So, the probability is 4/11.
Step 4: Multiply the probabilities together. The probability of both events happening is the product of the probabilities. So, the probability is 1/4 * 4/11 = 1/11.
So, the correct answer is B. StartFraction 1 over 11 EndFraction.
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