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There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probability
- a)P=1/3+(2/3*p^3)
- b)P=2/3+(2/3*p^3)
- c)P=2/3+(1/3*p^3)
- d)P=2/3+(2/3*p^3)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
There is a bacteria which has the probability of die 1/3 of its total ...
Probability of bacteria to die is 1/3
or its get tripled, hence the probability of getting tripled is (2/3)*p*p*p that is (2/3)*p^3
so both the probability combined as P=1/3+(2/3*p^3)
Most Upvoted Answer
There is a bacteria which has the probability of die 1/3 of its total ...
**Solution:**
Let's assume the initial number of bacteria is **p**.
According to the given information, there are two possible outcomes for each bacterium:
1. The bacterium dies with a probability of 1/3.
2. The bacterium triples with a probability of 2/3.
Now, let's calculate the probability for each outcome:
1. **The bacterium dies:**
The probability of the bacterium dying is 1/3. In this case, the total number of bacteria will decrease by 1.
So, the probability of this outcome is (1/3) * (p/p) = 1/3.
2. **The bacterium triples:**
The probability of the bacterium tripling is 2/3. In this case, the total number of bacteria will be multiplied by 3.
So, the probability of this outcome is (2/3) * (p/3p) = 2/9.
Now, let's calculate the overall probability using the Law of Total Probability:
P = P(die) + P(triple)
P = 1/3 + 2/9
P = 3/9 + 2/9
P = 5/9
Therefore, the correct answer is **option A: P = 1/3 * (2/3 * p^3)**.
This answer represents the overall probability of the given bacterium either dying or tripling its population.
Let's assume the initial number of bacteria is **p**.
According to the given information, there are two possible outcomes for each bacterium:
1. The bacterium dies with a probability of 1/3.
2. The bacterium triples with a probability of 2/3.
Now, let's calculate the probability for each outcome:
1. **The bacterium dies:**
The probability of the bacterium dying is 1/3. In this case, the total number of bacteria will decrease by 1.
So, the probability of this outcome is (1/3) * (p/p) = 1/3.
2. **The bacterium triples:**
The probability of the bacterium tripling is 2/3. In this case, the total number of bacteria will be multiplied by 3.
So, the probability of this outcome is (2/3) * (p/3p) = 2/9.
Now, let's calculate the overall probability using the Law of Total Probability:
P = P(die) + P(triple)
P = 1/3 + 2/9
P = 3/9 + 2/9
P = 5/9
Therefore, the correct answer is **option A: P = 1/3 * (2/3 * p^3)**.
This answer represents the overall probability of the given bacterium either dying or tripling its population.
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There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probabilitya)P=1/3+(2/3*p^3)b)P=2/3+(2/3*p^3)c)P=2/3+(1/3*p^3)d)P=2/3+(2/3*p^3)Correct answer is option 'A'. Can you explain this answer?
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