A point is chosen at random inside a rectangle measuring 6 inches by 5 inches. What is the probability that the randomly selected point is at least one inch from the edge of the rectangle?
This question asks for the probability of a point randomly selected inside a rectangle meeting a certain condition related to its distance from the edges. Probability in geometric problems like this is often calculated as the ratio of a favorable area to the total area.
The rectangle measures 6 inches by 5 inches. The total area of the rectangle is simply the product of its length and width.
Total Area = Length $\times$ Width
Total Area = \(6 \text{ inches} \times 5 \text{ inches}\)
Total Area = \(30 \text{ square inches}\)
This total area represents the entire sample space for where the random point can be located.
We are interested in the probability that the randomly selected point is at least one inch away from any edge of the rectangle. This means the point must be at least one inch from the top edge, at least one inch from the bottom edge, at least one inch from the left edge, and at least one inch from the right edge.
Imagine removing a 1-inch strip from each side of the original rectangle. The points that are at least one inch from all edges must lie within the boundaries of this inner region.
The favorable region where the point is at least one inch from all edges is a smaller rectangle with dimensions 4 inches by 3 inches.
Favorable Area = Inner Length $\times$ Inner Width
Favorable Area = \(4 \text{ inches} \times 3 \text{ inches}\)
Favorable Area = \(12 \text{ square inches}\)
The probability that the randomly selected point is at least one inch from the edge is the ratio of the favorable area (the inner rectangle) to the total area (the original rectangle).
Probability = \(\frac{\text{Favorable Area}}{\text{Total Area}}\)
Probability = \(\frac{12 \text{ square inches}}{30 \text{ square inches}}\)
Now, we simplify the fraction:
Probability = \(\frac{12}{30}\)
Both 12 and 30 are divisible by 6.
Probability = \(\frac{12 \div 6}{30 \div 6} = \frac{2}{5}\)
| Region | Dimensions | Area (sq inches) |
|---|---|---|
| Total Rectangle | 6" x 5" | 30 |
| Favorable Inner Rectangle (at least 1" from edge) | 4" x 3" | 12 |
Thus, the probability that the randomly selected point is at least one inch from the edge of the rectangle is \(\frac{2}{5}\).
| Concept | Description | Formula Example (Area) |
|---|---|---|
| Geometric Probability | Probability based on ratios of geometric measures (length, area, volume). | \(P(\text{event}) = \frac{\text{Measure of Favorable Region}}{\text{Measure of Total Sample Space}}\) |
| Sample Space (Total Area) | The entire region where the random point can be located. | For a rectangle: \(L \times W\) |
| Favorable Region (Favorable Area) | The sub-region where the event of interest occurs. | Calculated based on the conditions given (e.g., distance from edges). |
Geometric probability problems can involve various shapes and conditions. Here are a few examples of how the favorable region might be defined:
In all cases, the core principle remains the same: calculate the area (or length, or volume) of the total possible space and the area (or length, or volume) of the space that satisfies the condition, then take the ratio.
Remember to carefully define the boundaries of the favorable region based on the problem's constraints, such as "at least one inch from the edge" or "within 2 cm of the center".
Let A, B and C be three mutually exclusive and exhaustive events associated with a random experiment. If P (B) = 1.5 P (A) and P (C) = 0.5 P (B), then P (A) is equal to
A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?
In throwing of two dice, the number of exhaustive events that ‘5’ will never appear on any one of the dice is
Two independent events A and B have \({\rm{P}}\left( {\rm{A}} \right) = \frac{1}{3}\) and \({\rm{P}}\left( {\rm{B}} \right) = \frac{3}{4}\) . What is the probability that exactly one of the two events A or B occurs?
A problem in statistics is given to three students A, B and C whose chances of solving it independently are \(\frac{1}{2},\frac{1}{3}\) and \(\frac{1}{4}\) respectively. The probability that the problem will be solved is
The probabilities that a student will solve Question A and Question B are 0.4 and 0.5 respectively. What is the probability that he solves at least one of the two questions?
If a coin is tossed till the first head appears, then what will be the sample space?
A committee of three has to be chosen form a group of 4 men and 5 women. If the selection is made at random, what is the probability that exactly two members are men?
A pair of fair dice is rolled. What is the probability that the second dice lands on a higher value than does the first?
What is the probability of getting a composite number in the list of natural numbers from 1 to 50?
Let A, B and C be three mutually exclusive and exhaustive events associated with a random experiment. If P (B) = 1.5 P (A) and P (C) = 0.5 P (B), then P (A) is equal to
Five persons A, B, C, D and E occupy seats in a row at random. The probability that A and B sit next to each other is:
The probability of getting 9 cards of the same suit in one hand at a game of bridge is:
A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?
If the probability of A to fail in an examination is 0.2 and that for B is 0.3, then, the probability that either A or B fails is: