Find the probability that they speak the same fact if A speaks truth in 60% cases and B speaks truth in 75% cases?
Probability of Speaking the Same Fact: A Detailed Explanation
This problem asks us to find the probability that two individuals, A and B, speak the same fact, given their individual truth-telling percentages. Understanding the probabilities of them speaking the truth and lying is crucial to solving this problem.
Understanding the Given Probabilities
We are provided with the following information:
- Individual A speaks the truth in 60% cases.
- Individual B speaks the truth in 75% cases.
Let's define the probabilities for each individual:
- Probability that A speaks the truth, denoted as \(P(T_A)\): \(P(T_A) = 60\% = \frac{60}{100} = 0.60\)
- Probability that A lies (does not speak the truth), denoted as \(P(L_A)\): Since A either speaks the truth or lies, \(P(L_A) = 1 - P(T_A)\). \(P(L_A) = 1 - 0.60 = 0.40\)
- Probability that B speaks the truth, denoted as \(P(T_B)\): \(P(T_B) = 75\% = \frac{75}{100} = 0.75\)
- Probability that B lies (does not speak the truth), denoted as \(P(L_B)\): Since B either speaks the truth or lies, \(P(L_B) = 1 - P(T_B)\). \(P(L_B) = 1 - 0.75 = 0.25\)
Conditions for Speaking the Same Fact
For A and B to speak the same fact, there are two possible scenarios:
- Both speak the truth: This means A speaks the truth AND B speaks the truth.
- Both lie: This means A lies AND B lies.
Since the actions of A and B are independent (one person's truth-telling doesn't affect the other's), we can multiply their individual probabilities for combined events.
Calculating Probabilities for Each Scenario
Probability of Both Speaking Truth
The probability that both A and B speak the truth is the product of their individual probabilities of speaking the truth:
\[P(\text{A speaks truth AND B speaks truth}) = P(T_A) \times P(T_B)\] \[P(\text{Both Truth}) = 0.60 \times 0.75\] \[P(\text{Both Truth}) = 0.45\]Probability of Both Lying
The probability that both A and B lie is the product of their individual probabilities of lying:
\[P(\text{A lies AND B lies}) = P(L_A) \times P(L_B)\] \[P(\text{Both Lie}) = 0.40 \times 0.25\] \[P(\text{Both Lie}) = 0.10\]Finding the Overall Probability of Speaking the Same Fact
The total probability that they speak the same fact is the sum of the probabilities of these two mutually exclusive scenarios (both speak truth, or both lie):
\[P(\text{Speak Same Fact}) = P(\text{Both Truth}) + P(\text{Both Lie})\] \[P(\text{Speak Same Fact}) = 0.45 + 0.10\] \[P(\text{Speak Same Fact}) = 0.55\]To express this as a percentage, we multiply by 100:
\[P(\text{Speak Same Fact}) = 0.55 \times 100\% = 55\%\]Summary of Calculation
| Event | Probability | Calculation |
|---|---|---|
| A speaks truth | \(P(T_A) = 0.60\) | Given |
| A lies | \(P(L_A) = 0.40\) | \(1 - 0.60\) |
| B speaks truth | \(P(T_B) = 0.75\) | Given |
| B lies | \(P(L_B) = 0.25\) | \(1 - 0.75\) |
| Both speak truth | \(P(\text{Both Truth}) = 0.45\) | \(P(T_A) \times P(T_B) = 0.60 \times 0.75\) |
| Both lie | \(P(\text{Both Lie}) = 0.10\) | \(P(L_A) \times P(L_B) = 0.40 \times 0.25\) |
| Speak the same fact | \(P(\text{Same Fact}) = 0.55\) | \(P(\text{Both Truth}) + P(\text{Both Lie}) = 0.45 + 0.10\) |
Therefore, the probability that they speak the same fact is 55%.
Important Questions from Conditional Probability
A and B are two events such that A̅ and B̅ are mutually exclusive. If P(A) = 0.5 and P(B) = 0.6, then what is the value of P(A|B)?
For two dependent events A and B, it is given that P(A) = 0.2 and P(B) = 0.5. If A ⊆ B, then the values of conditional probabilities P(A|B) and P(B|A) are respectively
In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?
Let X1 and X2 be independent random variables each having geometric distribution qk p ; k = 0, 1, 2, …. Then the conditional distribution of X1 given X1 + X2 is
In a batch Of 100 resistors Of 1kiloohm resistance, 80 numbers are within the required tolerance values and II numbers are below the required tolerance values, the remaining are above the required tolerance values. If two resistors are drawn one after the other without replacement, the probability of the first one drawn is below and the second one drawn is above the required tolerance value is
