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
0.01
Probability Calculation for Resistor Selection
This problem involves calculating the probability of two dependent events occurring in sequence without replacement. We have a batch of resistors with different tolerance classifications, and we need to find the probability of drawing one resistor below tolerance, followed by one above tolerance.
Understanding the Resistor Batch Composition
First, let's break down the composition of the 100 resistors in the batch based on their tolerance values:
- Total resistors: 100
- Resistors within required tolerance values: 80 numbers
- Resistors below required tolerance values: 11 numbers
- Resistors above required tolerance values: The remaining numbers.
To find the number of resistors above the required tolerance values, we subtract the other categories from the total:
\( \text{Number of resistors above tolerance} = \text{Total resistors} - \text{Resistors within tolerance} - \text{Resistors below tolerance} \)
\( \text{Number of resistors above tolerance} = 100 - 80 - 11 = 9 \)
So, we have:
| Category | Number of Resistors |
|---|---|
| Within Tolerance | 80 |
| Below Tolerance | 11 |
| Above Tolerance | 9 |
| Total | 100 |
First Draw: Probability of Drawing a Resistor Below Tolerance
We want to find the probability that the first resistor drawn is below the required tolerance value. Initially, there are 100 resistors in total, and 11 of them are below tolerance.
\( P(\text{First resistor is below tolerance}) = \frac{\text{Number of resistors below tolerance}}{\text{Total number of resistors}} \)
\( P(\text{First below}) = \frac{11}{100} \)
Second Draw: Probability of Drawing a Resistor Above Tolerance (Without Replacement)
Since the first resistor drawn is not replaced, the total number of resistors for the second draw changes. After drawing one resistor (which was below tolerance), we are left with 99 resistors.
Crucially, the number of resistors above the required tolerance remains unchanged, as the first resistor drawn was from the 'below tolerance' category, not the 'above tolerance' category. So, there are still 9 resistors above tolerance.
\( P(\text{Second resistor is above tolerance | First was below tolerance}) = \frac{\text{Number of resistors above tolerance}}{\text{Remaining total number of resistors}} \)
\( P(\text{Second above | First below}) = \frac{9}{99} \)
We can simplify this fraction:
\( \frac{9}{99} = \frac{1 \times 9}{11 \times 9} = \frac{1}{11} \)
Combined Probability of Both Events
To find the probability of both events happening in this specific order (first is below tolerance AND second is above tolerance), we multiply the probabilities of the individual events. This is because these are dependent events.
\( P(\text{First below AND Second above}) = P(\text{First below}) \times P(\text{Second above | First below}) \)
Substitute the calculated probabilities:
\( P(\text{First below AND Second above}) = \frac{11}{100} \times \frac{9}{99} \)
Using the simplified fraction for the second probability:
\( P(\text{First below AND Second above}) = \frac{11}{100} \times \frac{1}{11} \)
Now, we can cancel out the '11' from the numerator and denominator:
\( P(\text{First below AND Second above}) = \frac{1}{100} \)
Converting this fraction to a decimal:
\( P(\text{First below AND Second above}) = 0.01 \)
Final Probability Result
The probability of the first resistor drawn being below the required tolerance and the second resistor drawn being above the required tolerance, when drawn one after the other without replacement, is 0.01.
Important Questions from Conditional Probability
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
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
