A box contains 25 parts of which 10 are defective. Two parts are being drawn simultaneously in a random manner from the box. The probability of both the parts being good is
7/20
Probability Calculation for Good Parts
This problem asks us to determine the probability of drawing two good parts when selecting them simultaneously from a box. The box contains a mix of both good and defective parts. To solve this, we will use the principles of combinations because the order in which the parts are drawn does not affect the outcome.
Parts Identification and Classification
Let's first identify the number of each type of part available in the box:
- Total number of parts in the box = 25
- Number of defective parts = 10
- Number of good parts = Total parts - Number of defective parts = 25 - 10 = 15 good parts
Total Possible Outcomes for Drawing Two Parts
To find the total number of ways to draw any two parts from the 25 available parts, we use the combination formula, which is given by \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). Here, \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose.
Total ways to draw 2 parts from 25 parts:
\( \text{Total Outcomes} = \binom{25}{2} = \frac{25 \times (25-1)}{2 \times 1} = \frac{25 \times 24}{2} \)
\( \text{Total Outcomes} = 25 \times 12 = 300 \)
Favorable Outcomes for Drawing Two Good Parts
Next, we need to find the number of ways to draw two good parts specifically from the 15 good parts identified earlier.
Ways to draw 2 good parts from 15 good parts:
\( \text{Favorable Outcomes} = \binom{15}{2} = \frac{15 \times (15-1)}{2 \times 1} = \frac{15 \times 14}{2} \)
\( \text{Favorable Outcomes} = 15 \times 7 = 105 \)
Calculating the Probability of Both Parts Being Good
The probability of an event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
\( \text{Probability (Both parts are good)} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} \)
\( \text{Probability (Both parts are good)} = \frac{105}{300} \)
Simplifying the Probability Fraction
To express the probability in its simplest form, we simplify the fraction \( \frac{105}{300} \). We can divide both the numerator and the denominator by common factors:
- First, divide both by 5: \( \frac{105 \div 5}{300 \div 5} = \frac{21}{60} \)
- Then, divide both by 3: \( \frac{21 \div 3}{60 \div 3} = \frac{7}{20} \)
Thus, the probability of both parts being good is \( \frac{7}{20} \).
Summary of Parts and Probability
| Description | Value |
|---|---|
| Total parts in the box | 25 |
| Defective parts | 10 |
| Good parts | 15 |
| Total ways to draw 2 parts | \( \binom{25}{2} = 300 \) |
| Ways to draw 2 good parts | \( \binom{15}{2} = 105 \) |
| Probability of both parts being good | \( \frac{105}{300} = \frac{7}{20} \) |
The final calculated probability that both the drawn parts are good is \( \frac{7}{20} \).
Important Questions from Introduction
In the following table, the left column contains the names of standard graph algorithms and the right column contains the time complexities of the algorithms. Here, n and m are number of vertices and edges, respectively. Match each algorithm with its time complexity.
List I List II Standard graph algorithms Time complexities A. Bellman‐Ford algorithm I. O(m*log n) B. Kruskal’s algorithm II. O(n 3) C. Floyd‐Warshall algorithm III. O(n*m) D. Topological sorting IV. O(n + m) Choose the correct answer from the options given below :
How many cards must be selected from a standard deck of 52 cards to guarantee that at least three hearts are present among them?
Match List 1 with List 2 and choose the correct answer from the code given below:
List I
(Graph Algorithm)
List II
(Time Complexity)
a) Dijkstra’s algorithm
i) Θ(E log E)
b) Kruskal’s algorithm
ii) Θ(V 3)
c) Floyd-Warshall algorithm
iii) Θ(V 2)
d) Topological sorting
iv) Θ(V + E)
Where V and E are the number of vertices and edges in graph respectively.
The solution of recurrence relation: T(n)=2T(sqrt(n)) + lg(n) is
Modulus of elasticity of concrete, E is calculated using:
