There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X > 3) is:
14/15
Understanding the Probability Problem
We are given 6 cards, numbered from 1 to 6. We draw two cards randomly without replacing the first card before drawing the second. We are interested in the sum of the numbers on these two cards, which we call X. Our goal is to find the probability that the sum X is greater than 3, denoted as P(X > 3).
Calculating the Total Number of Outcomes
First, let's determine the total number of possible ways to draw two cards from the six cards without replacement. Since the order in which we draw the two cards does not affect their sum, we can think of this as choosing a set of 2 cards from 6. This is a combination problem.
The total number of ways to choose 2 cards from 6 is given by the combination formula:
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
Here, $n=6$ (total number of cards) and $k=2$ (number of cards drawn).
$$ \text{Total outcomes} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} $$
$$ \binom{6}{2} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(4 \times 3 \times 2 \times 1)} = \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15 $$
So, there are 15 possible pairs of cards that can be drawn.
Finding the Probability P(X > 3)
We want to find P(X > 3), which is the probability that the sum of the numbers on the two cards is greater than 3.
The possible sums of two distinct numbers from the set {1, 2, 3, 4, 5, 6} range from 1+2=3 (minimum sum) to 5+6=11 (maximum sum). The sums X can take are 3, 4, 5, 6, 7, 8, 9, 10, 11.
Calculating the probability for each sum greater than 3 (X=4, 5, ..., 11) and adding them would be time-consuming. It's easier to use the concept of complementary probability.
The complementary event to X > 3 is X $\le$ 3. The only possible sum less than or equal to 3 is X = 3, as the minimum sum is 3.
So, P(X > 3) = 1 - P(X $\le$ 3) = 1 - P(X = 3).
Calculating P(X = 3)
Now, let's find the probability that the sum of the two drawn cards is exactly 3. We need to identify which pair(s) of distinct numbers from {1, 2, 3, 4, 5, 6} add up to 3.
The only pair of distinct numbers from the set {1, 2, 3, 4, 5, 6} that sums to 3 is {1, 2}.
There is only 1 favorable outcome where the sum X is 3.
The probability P(X = 3) is the number of favorable outcomes for X=3 divided by the total number of outcomes:
$$ P(X = 3) = \frac{\text{Number of pairs summing to 3}}{\text{Total number of pairs}} = \frac{1}{15} $$
Calculating P(X > 3) using Complementary Probability
Finally, we use the formula P(X > 3) = 1 - P(X = 3):
$$ P(X > 3) = 1 - \frac{1}{15} $$
To subtract the fractions, we find a common denominator:
$$ P(X > 3) = \frac{15}{15} - \frac{1}{15} = \frac{15 - 1}{15} = \frac{14}{15} $$
Thus, the probability that the sum of the numbers on the two cards drawn is greater than 3 is 14/15.
| Sum (X) | Pairs (without replacement) | Number of Pairs |
|---|---|---|
| 3 | {1, 2} | 1 |
| 4 | {1, 3} | 1 |
| 5 | {1, 4}, {2, 3} | 2 |
| 6 | {1, 5}, {2, 4} | 2 |
| 7 | {1, 6}, {2, 5}, {3, 4} | 3 |
| 8 | {2, 6}, {3, 5} | 2 |
| 9 | {3, 6}, {4, 5} | 2 |
| 10 | {4, 6} | 1 |
| 11 | {5, 6} | 1 |
| Total | 15 |
From the table, the number of outcomes for X > 3 is $1 + 2 + 2 + 3 + 2 + 2 + 1 + 1 = 14$. The probability is $14/15$. This matches the result obtained using the complementary probability method, confirming our answer.
Revision Table: Probability Concepts
| Concept | Description | Formula/Notation |
|---|---|---|
| Probability | The likelihood of an event occurring. | P(Event) = (Number of favorable outcomes) / (Total number of outcomes) |
| Combinations | The number of ways to choose a subset of items from a larger set where the order does not matter. | $$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$ |
| Drawing without Replacement | Once an item is drawn, it is not returned to the set, affecting subsequent draws. | Uses combinations or permutations depending on whether order matters. |
| Complementary Event | The event that a specific event does not occur. | P(Event') = 1 - P(Event) |
Additional Information: Sample Space Visualization
When dealing with drawing cards without replacement, it can be helpful to list or visualize the sample space. The sample space consists of all possible pairs of cards that can be drawn.
The 15 possible unique pairs are:
- {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}
- {2, 3}, {2, 4}, {2, 5}, {2, 6}
- {3, 4}, {3, 5}, {3, 6}
- {4, 5}, {4, 6}
- {5, 6}
We can then find the sum for each pair:
- Sums of 3: {1, 2} (1 pair)
- Sums of 4: {1, 3} (1 pair)
- Sums of 5: {1, 4}, {2, 3} (2 pairs)
- Sums of 6: {1, 5}, {2, 4} (2 pairs)
- Sums of 7: {1, 6}, {2, 5}, {3, 4} (3 pairs)
- Sums of 8: {2, 6}, {3, 5} (2 pairs)
- Sums of 9: {3, 6}, {4, 5} (2 pairs)
- Sums of 10: {4, 6} (1 pair)
- Sums of 11: {5, 6} (1 pair)
Summing the counts: 1 + 1 + 2 + 2 + 3 + 2 + 2 + 1 + 1 = 15, which matches our total number of outcomes.
The outcomes where the sum X > 3 are those where X = 4, 5, 6, 7, 8, 9, 10, or 11. The number of such outcomes is 1 + 2 + 2 + 3 + 2 + 2 + 1 + 1 = 14.
The probability P(X > 3) is then the number of favorable outcomes (14) divided by the total number of outcomes (15), which is 14/15.
Important Questions from Determinants
An even number is the determinant of which of the following matrices?
(A) \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\)
(B) \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\)
(C) \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\)
(D) \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\)
Choose the correct answer from the options given below:
Two pipes A and B together can fill a tank in 40 minutes. Pipe A is twice as fast as pipe B. Pipe A alone can fill the tank in :
A random variable X has the following probability distribution:
X | -2 | -1 | 0 | 1 | 2 -------------------------------------------- P(X) | 0.2 | 0.1 | 0.3 | 0.2 | 0.2
The variance of X will be:
The angle between two lines whose direction ratios are proportional to \( (\sqrt{3} - 1) \), \( (-\sqrt{3} - 1) \), and -4 is:
Given the determinant:
\[ \Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix} \]
Which of the following statements are correct?
(A) \( \Delta = 2(1 - \cos^2 x) \)
(B) \( \Delta = 2(2 - \sin^2 x) \)
(C) Minimum value of \( \Delta \) is 2
(D) Maximum value of \( \Delta \) is 4
Choose the correct answer from the options given below:
