A bag contains 4 black balls and 6 red balls, if one ball is drawn at random, then the probability of getting a red ball is
- \(\frac{5}{8}\)
- \(\frac{3}{5}\)
- \(\frac{1}{2}\)
- \(\frac{1}{56}\)
The Correct Option is B
Solution and Explanation
Correct answer: \(\frac{3}{5}\)
Explanation:
The total number of balls in the bag is: \[ 4 \, \text{(black balls)} + 6 \, \text{(red balls)} = 10 \, \text{balls} \] The number of red balls is 6. Therefore, the probability of drawing a red ball is: \[ P(\text{red ball}) = \frac{\text{number of red balls}}{\text{total number of balls}} = \frac{6}{10} = \frac{3}{5} \]
Hence, the probability of drawing a red ball is \(\frac{3}{5}\).
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:
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