Ten glass vases were to be packed one each in 10 boxes marked "Glass". Twelve brass vases were to be packed one each in 12 boxes marked "Brass". Four vases and boxes got mixed up. A customer orders 1 glass and 1 brass vase and is sent appropriately marked boxes. The chance that the customer does not get the ordered vases in correctly marked boxes is
1/3
Probability of Incorrect Vases in Boxes
This problem asks for the probability that a customer receiving a "Glass" box and a "Brass" box does not get the correct type of vase in the box, given a specific mix-up scenario.
Initially, we have 10 Glass vases to be packed in 10 boxes marked "Glass" and 12 Brass vases to be packed in 12 boxes marked "Brass".
The problem states that "Four vases and boxes got mixed up". We interpret this to mean that after the mix-up, exactly 4 vases are placed in the wrong type of box. Let \(N_{Gv\_Bb}\) be the number of Glass vases in Brass boxes and \(N_{Bv\_Gb}\) be the number of Brass vases in Glass boxes. The total number of vases in the wrong type of box is \(N_{Gv\_Bb} + N_{Bv\_Gb} = 4\).
Let's analyze the distribution of vases in boxes after the mix-up:
- Total Glass vases = 10
- Total Brass vases = 12
- Total Glass boxes = 10
- Total Brass boxes = 12
The number of Glass vases in Glass boxes is \(10 - N_{Gv\_Bb}\). The number of Brass vases in Brass boxes is \(12 - N_{Bv\_Gb}\).
The total number of vases in the 10 Glass boxes must be 10: \( (10 - N_{Gv\_Bb}) + N_{Bv\_Gb} = 10 \). This implies \(N_{Bv\_Gb} = N_{Gv\_Bb}\).
Since \(N_{Gv\_Bb} + N_{Bv\_Gb} = 4\) and \(N_{Gv\_Bb} = N_{Bv\_Gb}\), we must have \(N_{Gv\_Bb} = 2\) and \(N_{Bv\_Gb} = 2\).
So, after the mix-up, the distribution of vases in boxes is:
- Glass boxes: 10 total
- Glass boxes containing Glass vases: \(10 - N_{Bv\_Gb} = 10 - 2 = 8\)
- Glass boxes containing Brass vases: \(N_{Bv\_Gb} = 2\)
- Brass boxes: 12 total
- Brass boxes containing Brass vases: \(12 - N_{Gv\_Bb} = 12 - 2 = 10\)
- Brass boxes containing Glass vases: \(N_{Gv\_Bb} = 2\)
Customer Order and Event Definition
The customer orders 1 glass vase and 1 brass vase and is sent appropriately marked boxes. This means the customer receives one box marked "Glass" and one box marked "Brass".
We want to find the chance that the customer does not get the ordered vases in correctly marked boxes. This means either the "Glass" box contains a Brass vase, or the "Brass" box contains a Glass vase, or both.
Let A be the event that the box marked "Glass" contains a Brass vase.
Let B be the event that the box marked "Brass" contains a Glass vase.
We are looking for the probability of the event A or B occurring, which is \(P(A \cup B)\).
Calculating Probabilities
The probability of event A (Glass box contains Brass vase) is the number of Glass boxes with Brass vases divided by the total number of Glass boxes.
\[P(A) = \frac{\text{Number of Glass boxes with Brass vases}}{\text{Total number of Glass boxes}} = \frac{2}{10} = \frac{1}{5}\]The probability of event B (Brass box contains Glass vase) is the number of Brass boxes with Glass vases divided by the total number of Brass boxes.
\[P(B) = \frac{\text{Number of Brass boxes with Glass vases}}{\text{Total number of Brass boxes}} = \frac{2}{12} = \frac{1}{6}\]Since the selection of the "Glass" box is independent of the selection of the "Brass" box, events A and B are independent.
Probability of Not Getting Ordered Vases Correctly
The probability that the customer does not get the ordered vases correctly is the probability that event A occurs or event B occurs, \(P(A \cup B)\). For independent events, this is given by the formula:
\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]Since A and B are independent, \(P(A \cap B) = P(A) \times P(B)\).
\[P(A \cup B) = P(A) + P(B) - P(A)P(B)\] \[P(A \cup B) = \frac{1}{5} + \frac{1}{6} - \left(\frac{1}{5} \times \frac{1}{6}\right)\] \[P(A \cup B) = \frac{1}{5} + \frac{1}{6} - \frac{1}{30}\]To add and subtract these fractions, we find a common denominator, which is 30.
\[P(A \cup B) = \frac{1 \times 6}{5 \times 6} + \frac{1 \times 5}{6 \times 5} - \frac{1}{30}\] \[P(A \cup B) = \frac{6}{30} + \frac{5}{30} - \frac{1}{30}\] \[P(A \cup B) = \frac{6 + 5 - 1}{30}\] \[P(A \cup B) = \frac{11 - 1}{30}\] \[P(A \cup B) = \frac{10}{30}\] \[P(A \cup B) = \frac{1}{3}\]Thus, the chance that the customer does not get the ordered vases in correctly marked boxes is \(1/3\).
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
