A fair die is tossed repeatedly until a...

MathematicsRandom Variables and Probability DistributionsJEE Advanced 2009Easy

Comprehension Passage

A fair die is tossed repeatedly until a six is obtained. Let

X

denote the number of tosses required.

Question 1:

The probability that $X = 3$ equals

(A)

25/216

(B)

25/36

(C)

5/36

(D)

125/216

Answer:A

Question 2:

The probability that $X \geq 3$ equals

(A)

125/216

(B)

25/36

(C)

5/36

(D)

25/216

Answer:B

Question 3:

The conditional probability that $X \geq 6$ given $X > 3$ equals

(A)

125/216

(B)

25/216

(C)

5/36

(D)

25/36

Answer:D

Visualized Solution (Hindi)

Understanding the Random Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Let $X$ be the number of tosses until the first six appears.
Probability of success (getting a 6): $p = \frac{1}{6}$
Probability of failure (not getting a 6): $q = 1 - p = \frac{5}{6}$
This follows a Geometric Distribution.

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (X = 3)

For $X = 3$ , the sequence of outcomes must be: (Not 6, Not 6, 6)
$P (X = 3) = P (Failure) \times P (Failure) \times P (Success)$
$P (X = 3) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (X \geq 3)

The event $X \geq 3$ means the first six occurs on the 3rd, 4th, 5th... toss.
This is equivalent to saying: The first 2 tosses were NOT sixes.
$P (X \geq 3) = P (First 2 are failures) = (\frac{5}{6})^{2}$
$P (X \geq 3) = \frac{25}{36}$

General Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (X \geq k)

In general, for a geometric distribution: $P (X \geq k) = q^{k - 1}$
This represents the probability that the first $k - 1$ trials are failures.
Also note: $P (X > k) = P (X \geq k + 1) = q^{k}$

Conditional Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (X \geq 6∣ X > 3)

Using the conditional probability formula: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$
Here, $A = {X \geq 6}$ and $B = {X > 3}$ . Since $A \subset B$ , $A \cap B = {X \geq 6}$ .
$P (X \geq 6∣ X > 3) = \frac{P ( X \geq 6 )}{P ( X > 3 )} = \frac{P ( X \geq 6 )}{P ( X \geq 4 )}$
Substituting the formula $q^{k - 1}$ : $\frac{( 5/6 ) ^{5}}{( 5/6 ) ^{3}} = (\frac{5}{6})^{2} = \frac{25}{36}$

The Memoryless Property

Key Takeaway: The Geometric Distribution is memoryless.
$P (X \geq m + n ∣ X > m) = P (X \geq n)$
In our case: $P (X \geq 3 + 3∣ X > 3) = P (X \geq 3) = \frac{25}{36}$
Next Challenge: How would this change if we were looking for the second six instead of the first?

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Comprehension Passage

The probability that $X = 3$ equals

The probability that $X \geq 3$ equals

The conditional probability that $X \geq 6$ given $X > 3$ equals

Visualized Solution (Hindi)

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General Formula for $P (X \geq k)$

Conditional Probability $P (X \geq 6∣ X > 3)$

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Comprehension Passage

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that X=3 equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that X≥3 equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The conditional probability that X≥6 given X>3 equals

Visualized Solution (Hindi)

Understanding the Random Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=3)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X≥3)

General Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X≥k)

Conditional Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X≥6∣X>3)

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that X=3 equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The conditional probability that X≥6 given X>3 equals

A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(A∪B) is

A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss equals

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is

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The probability that $X = 3$ equals

The probability that $X \geq 3$ equals

The conditional probability that $X \geq 6$ given $X > 3$ equals

Understanding the Random Variable $X$

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Calculating $P (X \geq 3)$

General Formula for $P (X \geq k)$

Conditional Probability $P (X \geq 6∣ X > 3)$

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The probability that $X \geq 3$ equals

The conditional probability that $X \geq 6$ given $X > 3$ equals

A die is thrown. Let $A$ be the event that the number obtained is greater than 3. Let $B$ be the event that the number obtained is less than 5. Then $P (A \cup B)$ is