Question

Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, where x>y, be 8 and 16 respectively. Two numbers are chosen from \{1, 2, 3, x-4, y, 5\ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:}

• A. \(\frac{4}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{2}{5}\)
• D. \(\frac{1}{3}\)

Hint:

For probability problems involving "at least", "at most", or complex conditions, always consider calculating the probability of the complementary event. It's often much simpler and less prone to counting errors. Subtracting this from 1 gives the desired probability.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a two-part problem. First, we use the given mean and variance of a set of 7 observations to find the values of two unknown observations, x and y. Second, we use these values to form a new set of numbers and calculate a probability related to drawing two numbers from it.
Step 2: Key Formula or Approach:
1. Mean (\(\bar{x}\)): \(\bar{x} = \frac{\sum x_i}{n}\)
2. Variance (\(\sigma^2\)): \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\)
3. Probability: \(P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). We can use complementary probability \(P(A) = 1 - P(A')\), where A' is the complement of event A.
Step 3: Detailed Explanation:
Part 1: Finding x and y
The 7 observations are \{2, 4, 10, x, 12, 14, y\}. The number of observations \(n=7\).
Mean \(\bar{x} = 8\).
\[ \frac{2+4+10+x+12+14+y}{7} = 8 \] \[ 42 + x + y = 56 \implies x+y=14 \quad \text{(Eq. 1)} \] Variance \(\sigma^2 = 16\).
\[ \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 16 \] \[ \frac{2^2+4^2+10^2+x^2+12^2+14^2+y^2}{7} - 8^2 = 16 \] \[ \frac{4+16+100+x^2+144+196+y^2}{7} - 64 = 16 \] \[ \frac{460+x^2+y^2}{7} = 80 \] \[ 460+x^2+y^2 = 560 \implies x^2+y^2=100 \quad \text{(Eq. 2)} \] Now we solve the system of equations: \(x+y=14\) and \(x^2+y^2=100\).
From (1), \(y=14-x\). Substitute into (2):
\[ x^2 + (14-x)^2 = 100 \implies x^2 + 196 - 28x + x^2 = 100 \] \[ 2x^2 - 28x + 96 = 0 \implies x^2 - 14x + 48 = 0 \] Factor the quadratic: \((x-6)(x-8) = 0\). So, \(x=6\) or \(x=8\).
If \(x=6\), then \(y=14-6=8\). If \(x=8\), then \(y=14-8=6\).
We are given the condition \(x>y\), so we must have \(x=8\) and \(y=6\).
Part 2: Calculating the probability
The new set of numbers is \(S = \{1, 2, 3, x-4, y, 5\}\).
Substitute the values of x and y: \(x-4 = 8-4=4\), \(y=6\).
So the set is \(S = \{1, 2, 3, 4, 6, 5\}\) or, in order, \(S = \{1, 2, 3, 4, 5, 6\}\).
The size of the set is \(N=6\).
We choose two numbers one after another without replacement.
Total number of ordered pairs = \(6 \times 5 = 30\).
Let A be the event that the smaller of the two chosen numbers is less than 4.
This means the smaller number can be 1, 2, or 3.
It is easier to calculate the complement event, A': The smaller of the two chosen numbers is NOT less than 4, meaning the smaller number is \(\ge 4\).
For this to happen, both chosen numbers must be from the subset \(S' = \{4, 5, 6\}\).
Number of ways to choose 2 numbers from \(S'\) without replacement = \(3 \times 2 = 6\).
These pairs are (4,5), (4,6), (5,4), (5,6), (6,4), (6,5). In all these cases, the smaller number is \(\ge 4\).
So, the probability of event A' is:
\[ P(A') = \frac{\text{Number of outcomes in A'}}{\text{Total outcomes}} = \frac{6}{30} = \frac{1}{5} \] The probability of event A is:
\[ P(A) = 1 - P(A') = 1 - \frac{1}{5} = \frac{4}{5} \] Step 4: Final Answer:
The required probability is \(\frac{4}{5}\).