Problem 12 - A Complete Resource Book in Mathematics for JEE Main

Question

Cards are drawn from a pack of 52 cards one by one. The probability that exactly 10 cards will be drawn before the first ace is (A) \(\frac{451}{884}\) (B) \(\frac{241}{1456}\) (C) \(\frac{164}{4165}\) (D) none of these

Step-by-Step Solution

Answer

Option (C) \(\frac{164}{4165}\) is correct.

Step 1: Understanding the Problem

We have a standard deck of 52 playing cards. We need to determine the probability that the first occurrence of an Ace is exactly on the 11th draw. This means the first ten cards drawn are not Aces.

Step 2: Calculate the Probability of Not Drawing an Ace

The deck contains 4 Aces out of 52 cards, leaving 48 non-Ace cards. The probability that the first card is not an Ace is \(\frac{48}{52}\), the second is \(\frac{47}{51}\), continuing until the 10th card, which is \(\frac{39}{43}\).

Step 3: Probability of Drawing Ten Non-Aces

We multiply the probabilities of not drawing an Ace in each of the 10 draws: \(\frac{48}{52} \times \frac{47}{51} \times \ldots \times \frac{39}{43}\).

Step 4: Probability of Drawing an Ace on 11th Draw

After 10 non-Ace cards have been drawn, there are 42 cards left (48 original non-Aces minus 10). The probability that the 11th card is an Ace is \(\frac{4}{42}\).

Step 5: Combine Probabilities

Combine the result of the probability of drawing 10 non-Aces and then an Ace: the product of the two probabilities calculated. Simplifying this product gives us the final probability of \(\frac{164}{4165}\).

Key Concepts

Drawing Cards

Picture yourself drawing a card from a standard deck of 52 playing cards. Each deck includes 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, making up a total of 52 cards.
When drawing cards, it's crucial to remember the order or sequence matters, especially when calculating probabilities. If you are drawing cards one by one, without replacing them back into the deck, the probabilities change with each draw.
The sequence of drawing cards is essential in probability calculations as every draw reduces the number of potential outcomes. This is known as 'sampling without replacement,' which makes each subsequent draw dependent on the previous draws.

Non-Ace Cards

In probability exercises involving a deck of cards, understanding how many cards meet your criteria is vital. In this case, we're interested in non-Ace cards.
A standard deck has 4 aces, one from each suit, leaving us with 48 non-Ace cards. When you are focused on drawing non-Ace cards first, the probability for each draw will change and depend on how many non-Ace cards are left in the deck.
As you draw each non-Ace card, the chances of drawing another non-Ace decrease slightly because there are fewer non-Ace cards left. For instance, the probability for the first non-Ace is \(\frac{48}{52}\), then the next non-Ace is \(\frac{47}{51}\) because one non-Ace is already out, and so on until the desired number of non-Aces have been drawn.

Ace Probability Calculation

Let's look at how you can calculate the probability of drawing an Ace at a specific position, say on the 11th draw. First, we draw 10 non-Ace cards. We have already established how to multiply the probabilities for these 10 draws: \(\frac{48}{52} \times \frac{47}{51} \times \ldots \times \frac{39}{43}\).
When exactly 10 cards are drawn and all are non-Ace, there are 42 cards left in the deck. With all the reductions in non-Ace cards, the probability that the next card (11th) is an Ace is \(\frac{4}{42}\), considering no Ace has been drawn so far.
Now, to determine the probability the 11th card is the first Ace, we must combine the probability of drawing 10 non-Ace cards with the probability of then drawing an Ace: multiplying them together provides the solution. This illustrates how sequential probabilities depend not only on the events themselves but also on the outcomes of prior events, culminating in a specific final probability.