Binomial Distribution Questions

Binomial distribution questions for Class 12 with solutions are provided here for practice. Before understanding the concept of the binomial distribution, let us understand some facts about binomial experiments. There are some sorts of experiments which have only two possible outcomes, either a “success” or a “failure” – these types of random experiments are called binomial experiments or “Bernoulli trials”. For example, the experiment of tossing a coin and getting a head.

Thus, in a probability distribution, binomial distribution denotes the success of a random variable X in an n trials binomial experiment. Following are the conditions to find binomial distribution:

• n is finite and defined.
• Each trial has only two possible outcomes: success and failure.
• The result of each trial is independent of other trials.
• The probability of success and failure remains the same in each trial.

Bernoulli’s Theorem for Binomial Distribution

Let there be ‘n’ binomial experiment trials and let the random variable X denote the success of these trials. If p is the probability of success and 1 – p = q is the probability of failure in each trial, then,

As P(X) is the term of the binomial expansion of (p + q)ⁿ, it is called the binomial distribution.

Note :

• Sum of all probabilities in the distribution sums up to 1
• Probability of success in all n trials is pⁿ
• Probability of failure in all n trials is (1 – p)ⁿ = qⁿ
• Probability of success in at least one trial = P(X ≥ 1) = 1 – P(X = 0) = 1 – qⁿ.
• Probability of at least r successes = P(X ≥ r) = ∑_kⁿC_k p^k q^{n – k} (k = r, r + 1,…, n)
• Probability of at most r successes = P(X ≤ r) = ∑_kⁿC_k p^k q^{n – k} (k = 0, 1, …, r)
• If in n trials, the experiment is repeated N times, the expected frequencies are N.P(r) for r = 0, 1, 2, 3, …, n.

Learn more about binomial distribution.

Binomial Distribution Questions with Solutions

Let us practice some important questions on binomial distribution in probability.

Question 1:

Find the binomial distribution of getting a six in three tosses of an unbiased dice.

Solution:

Let X be the random variable of getting six. Then X can be 0, 1, 2, 3.

Here, n = 3

p = Probability of getting a six in a toss = ⅙

q = Probability of not getting a six in a toss = 1 – ⅙ = ⅚

P(X = 0) = ⁿC_r p^r q^{(n – r)} = ³C₀ (⅙)⁰ (⅚)^{3 – 0}

= 1 × 1 × 125/216 = 125/216

P(X = 1) = ⁿC_r p^r q^{(n – r)} = ³C₁ (⅙)¹ (⅚)^{3 – 1}

= 3 × ⅙ × 25/36 = 25/72

P(X = 2) = ⁿC_r p^r q^{(n – r)} = ³C₂ (⅙)² (⅚)^{3 – 2}

= 3 × 1/36 × ⅚ = 5/72

P(X = 3) = ⁿC_r p^r q^{(n – r)} = ³C₃ (⅙)³ (⅚)^{3 – 3}

= 1 × 1/216 × 1 = 1/216

The required binomial distribution of X is:

Question 2:

Find the probability distribution of the number of doublets in four throws of a pair of dice.

Solution:

There are 36 total possible outcomes for a throw of dice, for which the following outcomes are the success of the experiment: {(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

p = probability of getting doublets = 6/36 = ⅙

q = probability of getting not getting doublets = 1 – ⅙ = ⅚

X: numbers of doublets, then X = 0, 1, 2, 3, and 4.

P(X = 0) = ⁿC_r p^r q^{(n – r)} = ⁴C₀ (⅙)⁰ (⅚)^{4 – 0}

= 1 × 1 × 625/1296

P(X = 1) = ⁿC_r p^r q^{(n – r)} = ⁴C₁ (⅙)¹ (⅚)^{4 – 1}

= 4 × ⅙ × 125/216 = 125/324

P(X = 2) = ⁿC_r p^r q^{(n – r)} = ⁴C₂ (⅙)² (⅚)^{4 – 2}

= 6 × 1/36 × 25/36

= 25/216

P(X = 3) = ⁿC_r p^r q^{(n – r)} = ⁴C₃ (⅙)³ (⅚)^{4 – 3}

= 4 × 1/216 × ⅚ = 20/1296

P(X = 4) = ⁿC_r p^r q^{(n – r)} = ⁴C₄ (⅙)⁴ (⅚)^{4 – 4}

= 1 × 1/1296 × 1 = 1/1296.

∴ The required probability distribution is:

Question 3:

Find the probability of getting at least 5 times head-on tossing an unbiased coin for 6 times by using the binomial distribution.

Solution:

p = P(getting an head in a single toss) = ½

q = P(not getting an head in a single toss) = ½

X = successfully getting a head

P(X ≥ 5) = P(getting at least 5 heads) = P(X = 5) + P(X = 6)

= ⁶C₅ (½)⁵ (½)^{(6 – 5)} + ⁶C₆ (½)⁶ (½)^{6 – 6}

= 6 × (½)⁶ + 1 × (½)⁶ = 7/24.

Hence, the probability of getting at least 5 heads is 7/24.

Question 4:

There are four fused bulbs in a lot of 10 good bulbs. If three bulbs are drawn at random with replacement, find the probability of distribution of the number of fused bulbs drawn.

Solution:

This is a problem of binomial distribution as the event of drawing a fused bulb is independent.

p = P(drawing a fused bulb) = 4/(10 + 4) = 2/7

q = P(drawing a bulb which is not fused) = 1 – 2/7 = 5/7

X = event of drawing a fused bulb

X can take up the values 0, 1, 2, 3

P(X = 0) = P(getting zero fused bulbs in all draws)

= ⁿC_r p^r q^{(n – r)}

= ³C₀ (2/7)⁰ (5/7)^{(3 – 0)}

= 1 × 1 × (125/343) = 125/343

P(X = 1) = P (getting one time fused bulb)

= ⁿC_r p^r q^{(n – r)}

= ³C₁ (2/7)¹ (5/7)^{(3 – 1)}

= 3 × (2/7) × (25/49) = 150/343

P(X = 2) = P(getting two times fused bulbs)

= ⁿC_r p^r q^{(n – r)}

= ³C₂ (2/7)² (5/7)^{(3 – 2)}

= 3 × 4/49 × (5/7) = 60/343

P(X = 3) = (P(getting three times fused bulb)

= ⁿC_r p^r q^{(n – r)}

= ³C₃ (2/7)³ (5/7)^{(3 – 3)}

= 1 × 8/343 × 1 = 8/343

The required probability distribution:

Also Read:

• Random Variables
• Bayes Theorem
• Mean and Variance of Random Variables

Question 5:

On average, every one out of 10 telephones is found busy. Six telephone numbers are selected at random. Find the probability that four of them will be busy.

Solution:

Let X: event of getting a busy phone number

p = P(probability of getting a phone number busy) = 1/10

q = P(probability of not getting a phone number busy) = 9/10

The required probability = P(X = 4) = ⁶C₄ p⁴ q^{(6 – 4)}

= 15 × (1/10)⁴ × (9/10)²

= 15 × 81/10⁶

= 0.001215.

Question 6:

An unbiased dice is thrown until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw.

Solution:

Since each throw is independent of the previous throws, we can apply the binomial distribution formula to find the probability.

p = P(getting a six in a throw) = ⅙

q = P(not getting a six in a throw) = 1 – ⅙ = ⅚

According to the question, two sixes are already obtained in the previous throws.

∴ Required probability = P(getting exactly two sixes in five throws) × P(getting a six in the sixth throw)

= ⁵C₂ p²q³ × ¹C₁p¹ q ^{1 – 1}

= 10 × (⅙)² × (⅚)³ × 1 × (⅙)

= 10 × (⅙)³ × (⅚)³

= 625/23328.

Question 7:

The probability of a boy guessing a correct answer is ¼. How many questions must he answer so that the probability of guessing the correct answer at least once is greater than ⅔?

Solution:

p = P(guessing a correct answer) = ¼

q = P(not guessing a correct answer) = ¾

Let him answers n number of questions, then

P(X ≥ 1) = P(guessing at least one correct answer out of n questions) = 1 – P(no success) = 1 – qⁿ

Given, 1 – qⁿ > ⅔ ⇒ 1 – (¾)ⁿ > ⅔

⇒ (¾)ⁿ < ⅓

Now, let us check the above inequality for different values of n = 1, 2, 3, 4, …

When n = 1

¾ ≮ ⅓

When n = 2

(¾)² ≮ ⅓

When n = 3

(¾)³ ≮ ⅓

When n = 4

(¾)⁴ < ⅓.

Thus, he must answer at least 4 questions.

Question 8:

When a biased coin is tossed, the probability of getting a head 3 times more than the probability of getting a tail. Find the probability distribution for getting a tail, if the coin is tossed twice.

Solution:

Let the probability of getting a tail be p, then the probability of getting a head will be 3p

Now, p + 3p = 1 ⇒ p = ¼

q = P(not getting a tail) = 1 – ¼ = ¾

X = event of getting a tail in a toss

Then, possible values of x will be 0, 1, 2

P(X = 0) = ²C₀ p⁰ q^{2 – 0}

= 1 × 1 × (¾)²

= 9/16

P(X = 1) = ²C₁ p¹ q^{2 – 1}

= 2 × (¼) × (¾)

= ⅜

P(X = 2) = ²C₂ p² q^{2 – 2}

= 1 × (¼)² × 1

= 1/16

The probability distribution for getting the tail is:

Also try:Binomial Distribution Calculator

Question 9:

A bag contains 5 green balls and 3 red balls. If two balls are drawn from the bag randomly with replacement, find the probability distribution of the number of green balls drawn.

Solution:

Let p = P(getting a green ball) = 5/(5 + 3) = 5/8

q = P(not getting a green ball) = 1 – 5/8 = 3/8

X = event of drawing the green ball, then the value of X could be 0, 1, 2

P(X = 0) = Probability of getting no green ball = ²C₀ p⁰ q^{2 – 0} = 1 × 1 × (3/8)² = 9/64

P(X = 1) = Probability of getting one green ball = ²C₁ p¹ q^{2 – 1} = 2 × (⅝) × (⅜) = 15/32

P(X = 2) = Probability of getting 2 green balls = ²C₂ p² q^{2 – 2} = 1 × (⅝)² × (⅜)⁰ = 25/64

The required probability distribution is:

Question 10:

Find the probability distribution of getting the number of fours in three throws of a dice. Also, find the mean and variance of the distribution.

Solution:

Let, p = P(getting a four in a throw of dice) = ⅙

q = P(not getting a four in a throw of dice) = ⅚

X: number of four obtained, then the value of X could be 0, 1, 2, 3.

P(X = 0) = ³C₀ p⁰q^{3 – 0} = 1 × (⅚)³ = 125/216

P(X = 1) = ³C₁ p¹q^{3 – 1} = 3 × (⅙) × (⅚)² = 75/216

P(X = 2) = ³C₂ p²q^{3 – 2} = 1 × (⅙)² × (⅚)^{3 – 2} = 15/216

P(X = 3) = ³C₃ p³q^{3 – 3} = 1 × (⅙)³ × (⅚)^{3 – 3} = 1/216

The required probability distribution

Mean = np = 3 × ⅙ = 1.2

Variance = npq = 3 × ⅙ × ⅚ = 5/12.

Practice Problems on Binomial Distribution

1. Find the binomial distribution of getting an even number if an unbiased dice is thrown thrice.

2. How many times must a man toss an unbiased coin to get at least one head is more than 90%?

3. Three cards are drawn with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces drawn. Also, find the mean and variance of the distribution.

4. The probability of a trainee archer hitting the target is ¼. If he takes 7 shots, what is the probability of his hitting the target at least twice?

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