
The probability of drawing two red balls in succession from a bag containing 4 red and 5 black balls, when the ball that is drawn first is not replaced is
\[\begin{align}
& A.\dfrac{11}{81} \\
& B.\dfrac{4}{9} \\
& C.\dfrac{1}{6} \\
& D.\dfrac{1}{3} \\
\end{align}\]
Answer
557.7k+ views
Hint: In this question, we are given a number of red balls and black balls in a bag. If two balls are picked in succession without replacement, we need to find the probability of getting both red balls. For this, we will first find the probability of getting a red ball out of total balls using formula $\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total number of outcomes}}$.
Then we will find the probability of getting red balls from remaining red balls and remaining total balls. Multiplying both probabilities will give us our final answer.
Complete step by step answer:
Here we are given, number of red balls in the bag as 4 and number of black balls in the bag as 5. Two balls are picked in succession without replacement. So, we have to find the probability of getting both the balls as red.
Let us first calculate the probability of picking a red ball from the bag.
Total balls in the bag = Number of red balls + Number of black balls.
Total balls in the bag $\Rightarrow 4+5=9$.
As we know, $\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total number of outcomes}}$.
So for probability for picking red balls out of a ball, we get:
Number of favorable outcomes = 4.
Total number of outcomes = 9.
Hence, $\text{Probability}=\dfrac{\text{4}}{\text{9}}$.
Now, the ball is not replaced, so we are left with a total number of balls as 9-1 = 8 and number of red balls as 4-1 = 3.
For finding probability of getting red balls second time also, we have a favorable outcome as 3 and total number of outcomes as 8. So,
$\text{Probability}=\dfrac{\text{3}}{\text{8}}$.
Now, to find the probability of drawing two red balls in succession from a bag, we need to multiply both the probability.
Hence, we get required probability as $\dfrac{4}{9}\times \dfrac{3}{8}=\dfrac{12}{72}=\dfrac{1}{6}$.
So, the correct answer is “Option C”.
Note: Students should note that we have multiplied the probabilities because we need to find probability of occurring both events and not just one out of them. While calculating second probability, students can forget to decrease a red ball from total red balls or decrease a ball from total balls. Make sure that probability always lies between 0 and 1.
Then we will find the probability of getting red balls from remaining red balls and remaining total balls. Multiplying both probabilities will give us our final answer.
Complete step by step answer:
Here we are given, number of red balls in the bag as 4 and number of black balls in the bag as 5. Two balls are picked in succession without replacement. So, we have to find the probability of getting both the balls as red.
Let us first calculate the probability of picking a red ball from the bag.
Total balls in the bag = Number of red balls + Number of black balls.
Total balls in the bag $\Rightarrow 4+5=9$.
As we know, $\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total number of outcomes}}$.
So for probability for picking red balls out of a ball, we get:
Number of favorable outcomes = 4.
Total number of outcomes = 9.
Hence, $\text{Probability}=\dfrac{\text{4}}{\text{9}}$.
Now, the ball is not replaced, so we are left with a total number of balls as 9-1 = 8 and number of red balls as 4-1 = 3.
For finding probability of getting red balls second time also, we have a favorable outcome as 3 and total number of outcomes as 8. So,
$\text{Probability}=\dfrac{\text{3}}{\text{8}}$.
Now, to find the probability of drawing two red balls in succession from a bag, we need to multiply both the probability.
Hence, we get required probability as $\dfrac{4}{9}\times \dfrac{3}{8}=\dfrac{12}{72}=\dfrac{1}{6}$.
So, the correct answer is “Option C”.
Note: Students should note that we have multiplied the probabilities because we need to find probability of occurring both events and not just one out of them. While calculating second probability, students can forget to decrease a red ball from total red balls or decrease a ball from total balls. Make sure that probability always lies between 0 and 1.
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