Three cards were drawn from a pack of 52 cards. The probability that they are a king, a queen, and a jack is
Understanding Card Probability
This problem asks us to find the probability of drawing a specific set of three cards – a king, a queen, and a jack – from a standard pack of 52 cards. To solve this, we need to understand the concepts of combinations and probability.
Probability is calculated as the ratio of favorable outcomes to the total possible outcomes:
$$ \text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} $$
Total Possible Outcomes for Drawing Cards
First, let's determine the total number of ways to draw 3 cards from a pack of 52 cards. Since the order in which the cards are drawn does not matter, this is a combination problem. The formula for combinations is:
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
Where \(n\) is the total number of items to choose from, and \(k\) is the number of items to choose.
- Total cards in the pack, \(n = 52\)
- Number of cards to draw, \(k = 3\)
So, the total number of ways to draw 3 cards from 52 cards is:
$$ \text{Total Outcomes} = \binom{52}{3} = \frac{52!}{3!(52-3)!} $$
$$ \text{Total Outcomes} = \frac{52 \times 51 \times 50}{3 \times 2 \times 1} $$
Let's perform the calculation:
- \(52 \times 51 \times 50 = 132600\)
- \(3 \times 2 \times 1 = 6\)
- $$ \text{Total Outcomes} = \frac{132600}{6} = 22100 $$
Therefore, there are 22,100 total possible outcomes when drawing 3 cards from a 52-card deck.
Favorable Outcomes for Drawing Specific Cards
Next, we need to find the number of favorable outcomes, which is drawing one king, one queen, and one jack. A standard pack of 52 cards has:
- 4 kings
- 4 queens
- 4 jacks
We need to choose 1 king from 4, 1 queen from 4, and 1 jack from 4. These are independent selections.
- Number of ways to choose 1 king from 4 = \( \binom{4}{1} = 4 \)
- Number of ways to choose 1 queen from 4 = \( \binom{4}{1} = 4 \)
- Number of ways to choose 1 jack from 4 = \( \binom{4}{1} = 4 \)
To find the total number of favorable outcomes, we multiply these possibilities:
$$ \text{Favorable Outcomes} = \binom{4}{1} \times \binom{4}{1} \times \binom{4}{1} $$
$$ \text{Favorable Outcomes} = 4 \times 4 \times 4 = 64 $$
So, there are 64 favorable outcomes for drawing one king, one queen, and one jack.
Final Probability Calculation
Now, we can calculate the probability by dividing the favorable outcomes by the total possible outcomes:
$$ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} $$
$$ \text{Probability} = \frac{64}{22100} $$
To simplify this fraction, we can divide both the numerator and the denominator by their greatest common divisor. Both numbers are divisible by 4:
- \(64 \div 4 = 16\)
- \(22100 \div 4 = 5525\)
So, the simplified probability is:
$$ \text{Probability} = \frac{16}{5525} $$
Conclusion
The probability of drawing a king, a queen, and a jack when three cards are drawn from a pack of 52 cards is \( \frac{16}{5525} \).
| Description | Value |
|---|---|
| Total cards in a pack | 52 |
| Cards drawn | 3 |
| Total Possible Outcomes (Combinations) | $$ \binom{52}{3} = 22100 $$ |
| Number of Kings | 4 |
| Number of Queens | 4 |
| Number of Jacks | 4 |
| Favorable Outcomes (Choosing 1 King, 1 Queen, 1 Jack) | $$ \binom{4}{1} \times \binom{4}{1} \times \binom{4}{1} = 64 $$ |
| Probability | $$ \frac{64}{22100} = \frac{16}{5525} $$ |
Important Questions from Sampling Theorems
Four red balls, four green balls and four blue balls are put in a box. Three balls are pulled out of the box at random one after another without replacement. The probability that all the three balls are red is
A population (with mean $\mu$) follows normal distribution. Ten samples (N) are drawn at random with a mean value of “x” and standard deviation of “S”. Following table provides the confidence limits, C(t) of the cumulative probability function for Student's t - distribution two-tailed test with degree of freedom, D.
C(t)
D 0.9 0.95 0.975 9 1.38 1.83 2.26 10 1.37 1.81 2.23 11 1.36 1.80 2.20 Which one of the following expression is correct for testing the null hypothesis $H_0: \mu = 0$ at $10\%$ significance level?
- If the sample size ($n$) is 25 and the standard deviation ($\sigma$) of population is 2, then the standard error (SE) of sample mean, (rounded off to one decimal place), is ________.
The probability distribution function of a random variable $X$ is shown in the following figure.
From this distribution, random samples with sample size $n = 68$ are taken. If $\bar{X}$ is the sample mean, the standard deviation of the probability distribution of $\bar{X}$, i.e. $\sigma_{\bar{X}}$ is ________ (round off to 3 decimal places).
The standard deviation of population P is two times the standard deviation of population Q. The size of a random sample from population P is four times the size of a random sample from population Q. If $e_P$ and $e_Q$ denote the standard error of means of the samples from P and Q, respectively, then the ratio of $e_P$ to $e_Q$ is________.
