A bag contains 3 red balls and 5 black balls. If a ball is drawn at random from the bag, then the probability of getting a red ball is
- \(\frac{1}{2}\)
- \(\frac{3}{4}\)
- \(\frac{5}{8}\)
- \(\frac{3}{8}\)
The Correct Option is D
Approach Solution - 1
Given: A bag contains 3 red balls and 5 black balls.
Step 1: Understanding the Probability of Selecting a Red Ball
Total balls = 3 + 5 = 8
Red balls = 3
Step 2: Probability Calculation
Probability of selecting a red ball = Number of red balls / Total number of balls
= 3 / 8
Final Answer: 3 / 8
Approach Solution -2
The problem involves finding the probability of drawing a red ball from a bag containing 3 red balls and 5 black balls.
To determine the probability of an event, use the formula:
Probability of an event = (Number of favorable outcomes) / (Total number of possible outcomes)
In this case, the favorable outcome is drawing a red ball. T
herefore, the number of favorable outcomes is the number of red balls, which is 3.
The total number of possible outcomes is the total number of balls, which is the sum of red and black balls:
Total number of balls: 3 (red) + 5 (black) = 8
Now, apply the probability formula:
Probability of drawing a red ball = Number of red balls / Total number of balls = 3 / 8
Thus, the probability of drawing a red ball is \(\frac{3}{8}\).
The correct answer is \(\frac{3}{8}\).
Top Questions on Probability
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
(i) A person was tested randomly. What is the probability that he/she has contracted the disease?}
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category \( A_2 \)?- If A and B are two events such that $P(B) = \frac{1}{5}$, $P(A | B) = \frac{2}{3}$ and $P(A \cup B) = \frac{3}{5}$, then $P(A)$ is :
- If \( P(A) = \frac{1}{5}, P(B) = \frac{3}{5} \) and \( P(A \cap B) = \frac{2}{5} \), then \( P(A' \cup B') \) is :
- Bag I contains 4 white and 5 black balls. Bag II contains 6 white and 7 black balls. A ball drawn randomly from Bag I is transferred to Bag II and then a ball is drawn randomly from Bag II. Find the probability that the ball drawn is white.
- A meeting will be held only if all three members A, B and C are present. The probability that member A does not turn up is 0.10, member B does not turn up is 0.20 and member C does not turn up is 0.05. The probability of the meeting being cancelled is:
Questions Asked in AP POLYCET exam
- Area of a sector of a circle with radius 4 cm and angle 30° is (use \( \pi = 3.14 \)):
- The solution of \( x - 2y = 0 \) and \( 3x + 4y - 20 = 0 \) is:
- AP POLYCET - 2025
- Linear Equations
- If assumed mean of a data is 47.5, \( \sum f_i d_i = 435 \) and \( \sum f_i = 30 \), then mean of that data is:
- AP POLYCET - 2025
- Arithmetic Mean
- The points \( (1,5), (2,3) \) and \( (-2, -11) \) form a:
- A prime number \( p \) divides \( a^2 \) where \( a \) is a positive integer, then
- AP POLYCET - 2025
- Prime and Composite Numbers