From two well shuffled pack of cards what is the probability of getting one Jack from the first one and a King from the second?
Understanding Probability with Cards
This problem asks for the probability of two specific events happening when drawing from two separate, well-shuffled decks of cards. The key is that the events are independent because the outcome of drawing from the first deck does not affect the outcome of drawing from the second deck.
Probability of Drawing from a Deck of Cards
A standard deck of cards has 52 cards. These 52 cards include:
- 4 suits (Hearts, Diamonds, Clubs, Spades)
- 13 ranks in each suit (2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace)
The number of Jacks in a standard deck is 4 (one for each suit). The number of Kings in a standard deck is also 4 (one for each suit).
The probability of drawing a specific type of card from a well-shuffled deck is calculated as:
\(\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)
Calculating Probability for Each Deck
We are dealing with two separate, well-shuffled packs of cards.
Probability from the First Pack (Getting a Jack)
For the first pack, we want to get a Jack.
- Number of favorable outcomes (getting a Jack): 4 (since there are 4 Jacks)
- Total number of possible outcomes (total cards in the deck): 52
So, the probability of getting a Jack from the first pack is:
\(P(\text{Jack from first pack}) = \frac{4}{52} = \frac{1}{13}\)
Probability from the Second Pack (Getting a King)
For the second pack, we want to get a King.
- Number of favorable outcomes (getting a King): 4 (since there are 4 Kings)
- Total number of possible outcomes (total cards in the deck): 52
So, the probability of getting a King from the second pack is:
\(P(\text{King from second pack}) = \frac{4}{52} = \frac{1}{13}\)
Combined Probability of Independent Events
Since the draws from the two decks are independent events, the probability of both events happening is the product of their individual probabilities.
\(P(\text{Jack from first AND King from second}) = P(\text{Jack from first}) \times P(\text{King from second})\)
Substitute the probabilities we calculated:
\(P(\text{Jack from first AND King from second}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{13 \times 13}\)
Summarizing the Probabilities
| Event | Number of Favorable Outcomes | Total Outcomes | Probability |
|---|---|---|---|
| Getting a Jack from the first pack | 4 | 52 | \(\frac{4}{52} = \frac{1}{13}\) |
| Getting a King from the second pack | 4 | 52 | \(\frac{4}{52} = \frac{1}{13}\) |
| Both events happening | - | - | \(\frac{1}{13} \times \frac{1}{13} = \frac{1}{13 \times 13}\) |
The calculated probability of getting one Jack from the first well-shuffled pack and a King from the second well-shuffled pack is \(\frac{1}{13 \times 13}\).
Important Questions from Probability of Random Experiments
A six faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
A coin is biased so that a head is twice as likely to occur as a tail, if the coin is tossed three times, what is the probability of getting exactly two tails?
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One summer, Peter visits 4 villages (A, B, C and D) in a random order. Find the probability that he visits (i) A before B (ii) A before B and B before C respectively?
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