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Animated Solution for Mathematics - Probability: A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then $n$ is equal to

Enter Numerical Value:

Visualized Solution

The Setup: A Lost Card

Total cards in a standard deck = $52$ .
One card is lost, leaving $51$ cards.
We draw $n$ cards from the remaining $51$ , and all are spades.

Defining the Hypotheses (Events $E_{1}$ and $E_{2}$ )

Let $E_{1}$ be the event: 'The lost card is a spade.'
Let $E_{2}$ be the event: 'The lost card is NOT a spade.'
Total spades = $13$ , Total non-spades = $39$ .

Calculating Prior Probabilities

$P (E_{1}) = \frac{13}{52} = \frac{1}{4}$
$P (E_{2}) = \frac{39}{52} = \frac{3}{4}$

The Observed Event $A$

Let $A$ be the event: 'Drawing $n$ spades from the remaining $51$ cards.'
We need to find the probability of $A$ under each hypothesis.

Conditional Probability $P (A ∣ E_{1})$

If $E_{1}$ occurs, $1$ spade is lost.
Remaining spades = $12$ . Remaining cards = $51$ .
$P (A ∣ E_{1}) = \frac{^{12} C _{n}}{^{51} C _{n}}$

Conditional Probability $P (A ∣ E_{2})$

If $E_{2}$ occurs, a non-spade is lost.
Remaining spades = $13$ . Remaining cards = $51$ .
$P (A ∣ E_{2}) = \frac{^{13} C _{n}}{^{51} C _{n}}$

Applying Bayes' Theorem

We are given that the lost card is a spade, given $A$ occurred.
We need $P (E_{1} ∣ A) = \frac{11}{50}$ .
Formula: $P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} )}$

Substituting the Probabilities

$P (E_{1} ∣ A) = \frac{\frac{1}{4} \cdot \frac{^{12} C _{n}}{^{51} C _{n}}}{\frac{1}{4} \cdot \frac{^{12} C _{n}}{^{51} C _{n}} + \frac{3}{4} \cdot \frac{^{13} C _{n}}{^{51} C _{n}}}$

Simplifying the Expression

Cancel $\frac{1}{4}$ and $^{51} C_{n}$ from the numerator and denominator.
$P (E_{1} ∣ A) = \frac{^{12} C _{n}}{^{12} C _{n} + 3 \cdot ^{13} C _{n}}$

Expanding the Combinations Ratio

Divide numerator and denominator by $^{12} C_{n}$ .
We need the ratio $\frac{^{13} C _{n}}{^{12} C _{n}}$ .
$\frac{^{13} C _{n}}{^{12} C _{n}} = \frac{13}{13 - n}$

Finalizing the Probability Expression

Substitute the ratio back:
$P (E_{1} ∣ A) = \frac{1}{1 + 3 \cdot ( \frac{13}{13 - n} )}$
$P (E_{1} ∣ A) = \frac{13 - n}{( 13 - n ) + 39} = \frac{13 - n}{52 - n}$

Equating to the Given Value

We are given $P (E_{1} ∣ A) = \frac{11}{50}$ .
$\frac{13 - n}{52 - n} = \frac{11}{50}$

Solving the Linear Equation

Cross-multiply: $50 (13 - n) = 11 (52 - n)$
$650 - 50 n = 572 - 11 n$

Finding the Value of $n$

Rearranging terms: $650 - 572 = 50 n - 11 n$
$78 = 39 n$
$n = 2$

Final Answer

The number of spades drawn is $n = 2$ .
Key Takeaway: Bayes' theorem problems simplify significantly when common terms in combinations are factored out.

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The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

Imagine you are standing at a table with a standard deck of

52

cards. You know the composition perfectly:

13

spades and

39

non-spades.

Suddenly, a card slips from the deck and vanishes. You don't know what it is. You are then asked to draw

n

cards from the remaining

51

, and to your surprise, every single one of them is a spade.

The question is: what is the probability that the lost card was a spade? Furthermore, what is the value of

n

that makes this probability exactly

\frac{11}{50}

Defining the Hypotheses

We have two competing realities, which we call hypotheses:

E_{1}

: The lost card is a spade. -

E_{2}

: The lost card is NOT a spade.

Before we draw any cards, we determine the prior probabilities. Since there are

13

spades in a

52

-card deck, the probability of losing a spade is

P (E_{1}) = \frac{13}{52} = \frac{1}{4}

Consequently, the probability of losing a non-spade is

P (E_{2}) = \frac{39}{52} = \frac{3}{4}

The Observed Reality

Now, we observe Event

A

: "Drawing

n

spades from the remaining

51

cards." We calculate the conditional probability of

A

occurring under each hypothesis.

E_{1}

is true (a spade was lost), we have

12

spades left in

51

cards. The probability of drawing

n

spades is:

P (A ∣ E_{1}) = \frac{^{12} C _{n}}{^{51} C _{n}}

E_{2}

is true (a non-spade was lost), we still have

13

spades in

51

cards. The probability of drawing

n

spades is:

P (A ∣ E_{2}) = \frac{^{13} C _{n}}{^{51} C _{n}}

The Bayes' Bridge

We want to find

P (E_{1} ∣ A)

, the probability that the lost card was a spade given that we drew

n

spades. Bayes' Theorem provides the path:

P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} )}

Substituting our values into the equation:

P (E_{1} ∣ A) = \frac{\frac{1}{4} \cdot \frac{^{12} C _{n}}{^{51} C _{n}}}{\frac{1}{4} \cdot \frac{^{12} C _{n}}{^{51} C _{n}} + \frac{3}{4} \cdot \frac{^{13} C _{n}}{^{51} C _{n}}}

The Algebraic Dance

The

\frac{1}{4}

and the

^{51} C_{n}

appear in every term. We can divide them out entirely to simplify the expression:

P (E_{1} ∣ A) = \frac{^{12} C _{n}}{^{12} C _{n} + 3 \cdot ^{13} C _{n}}

To simplify further, divide the numerator and denominator by

^{12} C_{n}

P (E_{1} ∣ A) = \frac{1}{1 + 3 \cdot \frac{^{13} C _{n}}{^{12} C _{n}}}

Using the property

\frac{^{m} C _{r}}{^{m - 1} C _{r}} = \frac{m}{m - r}

, we find that

\frac{^{13} C _{n}}{^{12} C _{n}} = \frac{13}{13 - n}

. Substituting this back:

P (E_{1} ∣ A) = \frac{1}{1 + 3 \cdot ( \frac{13}{13 - n} )} = \frac{13 - n}{( 13 - n ) + 39} = \frac{13 - n}{52 - n}

The Final Reveal

We are given that

P (E_{1} ∣ A) = \frac{11}{50}

. We set up the equation:

\frac{13 - n}{52 - n} = \frac{11}{50}

Cross-multiplying gives:

50 (13 - n) = 11 (52 - n)

650 - 50 n = 572 - 11 n

78 = 39 n

Solving for

n

, we get:

$n = 2$

Common Traps & FAQs

Wait, why can't we just use simple probability instead of Bayes' Theorem?

Simple probability works when you are predicting a future outcome. Here, the outcome (drawing

n

spades) has already occurred. We are looking for the probability of a past event (the lost card) given this new information. This 'reverse' logic is exactly what Bayes' Theorem is designed for!

How do we handle the combinations

^{12} C_{n}

and

^{13} C_{n}

without getting lost in the algebra?

The secret is to not expand them immediately! Notice that both

P (A ∣ E_{1})

and

P (A ∣ E_{2})

have the same denominator,

^{51} C_{n}

. When you plug them into Bayes' formula, these denominators cancel out. Then, divide the remaining terms by

^{12} C_{n}

to simplify the ratio

^{13} C_{n} /^{12} C_{n}

into a simple fraction:

\frac{13}{13 - n}

Is there a physical intuition for why the probability changes?

Absolutely! Think of it this way: if the lost card was a spade, the deck is 'poorer' in spades, making it harder to draw

n

spades. If the lost card was not a spade, the deck is 'richer' in spades, making it easier. The observed event (drawing

n

spades) acts as evidence that shifts our belief about which card was lost.

Animated Solution for Mathematics - Probability: A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then $n$ is equal to

Enter Numerical Value:

Visualized Solution

The Setup: A Lost Card

Defining the Hypotheses (Events $E_{1}$ and $E_{2}$ )

Calculating Prior Probabilities

The Observed Event $A$

Conditional Probability $P (A ∣ E_{1})$

Conditional Probability $P (A ∣ E_{2})$

Applying Bayes' Theorem

Substituting the Probabilities

Simplifying the Expression

Expanding the Combinations Ratio

Finalizing the Probability Expression

Equating to the Given Value

Solving the Linear Equation

Finding the Value of $n$

Final Answer

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Hypotheses

The Observed Reality

The Bayes' Bridge

The Algebraic Dance

The Final Reveal

Similar Questions

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$ , then $n$ is equal to ____.

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $n^{2} - m^{2}$ is equal to :

Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box 1 is :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Comprehension Passage

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Enter Numerical Value:

Visualized Solution

The Setup: A Lost Card

Defining the Hypotheses (Events .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​)

Calculating Prior Probabilities

The Observed Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Conditional Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A∣E1​)

Conditional Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A∣E2​)

Applying Bayes' Theorem

Substituting the Probabilities

Simplifying the Expression

Expanding the Combinations Ratio

Finalizing the Probability Expression

Equating to the Given Value

Solving the Linear Equation

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Answer

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Hypotheses

The Observed Reality

The Bayes' Bridge

The Algebraic Dance

The Final Reveal

Similar Questions

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box 1 is :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Comprehension Passage

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Defining the Hypotheses (Events $E_{1}$ and $E_{2}$ )

The Observed Event $A$

Conditional Probability $P (A ∣ E_{1})$

Conditional Probability $P (A ∣ E_{2})$

Finding the Value of $n$