A coin is tossed until a head appears or

CUET Preparation Today

Start Free Mocks

Home Questions XX4JMTR75F

Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

A coin is tossed until a head appears or the tail appears 4 times in succession. Find the probability distribution of the number of tosses.

Options:

$X$	0	1	2	3
$P(X)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

$X$	0	1	2	3
$P(X)$	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{8}$

$X$	0	1	2	3
$P(X)$	$\frac{1}{2}$	$\frac{3}{4}$	$\frac{3}{8}$	$\frac{1}{8}$

$X$	0	1	2	3
$P(X)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{4}$	$\frac{1}{2}$

Correct Answer:

$X$	0	1	2	3
$P(X)$	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{8}$

Explanation:

The correct answer is Option (2) →

$X$	0	1	2	3
$P(X)$	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{8}$

The sample space associated with this experiment is

$S = \{H, TH, TTH, TTTH, TTTT\}$

Let the random variable X be defined as the number of tosses. Then X can take the values 1, 2, 3, 4. The corresponding probabilities are

$P(X = 1) = P(H)=\frac{1}{2}$

$P(X = 2) = P(TH)=\frac{1}{4}$

$P(X = 3)= P(TTH)=\frac{1}{8}$

$P(X = 4) = P(TTTH, TTTT)=\frac{1}{8}$

We observe that $ΣP(X)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=1$

The probability distribution of the number of tosses is

$X$	0	1	2	3
$P(X)$	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{8}$