Binomial Distribution Test - 1

Concept:

The binomial distribution:

The binomial distribution is a discrete probability distribution of the successes in a sequence of n independent yes/no experiments.

Calculation:

The binomial distribution is a discrete probability distribution of the successes in a sequence of n independent yes/no experiments. 

The binomial distribution has typically two parameters first is n, number of trials and other is the probability of success 'p'.

Concept:

• Mean = μ = np
• \({\rm{Standard\;deviation}} = {\rm{\sigma }} = \sqrt {{\rm{npq}}} {\rm{\;}}\)
• Variance = npq

Where n = number of trials; p = probability of success; q = (1-p) = probability of failure

probability of an event lies between 0 and 1.

Calculation:

As we know, Mean = μ = np

Variance = npq 

⇒ Variance = μ × q

Probability of an event lies between 0 and 1.

Therefore 0 ≤ {p, q} ≤ 1

μ × q < μ

∴ Mean > Variance

Given 0

Mean of binomial distribution = np = 4

Probability of success (p) = 0.4

Variance of binomial distribution = npq

Calculation

p = 0.4 and np = 4

⇒ 0.4n = 4

⇒ n = 10

⇒ p = 1 – q

⇒ q = 1 – 0.4 = 0.6

∴ Variance of binomial distribution = 10 × 0.4 × 0.6

= 2.4

Concept:

Since both the positive and negative values are likely to occur, the negative values are binomially distributed

Binomial distribution:

 If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = nCk pk (1 - p)(n - k)

Where,

n is the number of observations, k is the number of success

p is the probability of success & (1 - p) is the probability of failure.

Calculation:

Given:

n = 5, p = 1/2

Let X represents number of negative values in 5 trials

At most 1 negative value means, k can be 0 and 1

P(At most one negative value) = P(X≤1)

P(X ≤ 1) = P(X = 0) + P(X = 1)

\({\rm{P}}\left( {{\rm{X\;}} = {\rm{\;}}0} \right) = 5{{\rm{C}}_0\times }{\left( {\frac{1}{2}} \right)^0} \times {\left( {\frac{1}{2}} \right)^5=1/32}\)

\({\rm{P}}\left( {{\rm{X\;}} = {\rm{\;}}1} \right) = 5{{\rm{C}}_1\times }{\left( {\frac{1}{2}} \right)^1} \times {\left( {\frac{1}{2}} \right)^4=5/32}\)

⇒ 6/32

Important Points

• At most mean maximum. This term usually used when there are multiple trials like tossing a coin, throwing a dice
• For example in tossing two coins
• At most two head means it can be two head or one head or no head

Concept:

Probability of exactly r success in n trials = p(x = r) = ⁿC_r p^r q^{(n  -r)}  

Where n = number of trials; p = probability of success; q = (1-p) = probability of failure

Calculation:

As we know that, p(x = r) = ⁿC_r p^r q^{(n - r)}  

Given: 3P(X = 2) = P(X = 3) and n = 5

⇒ 3 × ⁵C₂ p² q^{(5 - 2)} = ⁵C₃ p³ q^{(5 - 3)}  

⇒ 3 × 10 × p² × q³ = 10 × p³ × q²

⇒ 3q = p

⇒ 3 (1 – p) = p

⇒ 3 – 3p = p

⇒ 4p = 3

∴ p = 3/4

Concept:

Binomial distribution:

If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = nCk pk q(n - k) where q = p - 1, n is the number of observations, p is the probability of success & q is the probability of failure

Note: The mean of a binomial distribution is np and variance is npq.

Calculation:

Given: An unbiased die is thrown 72 times and getting a 4 on a throw is termed as success

As we know that when a die is thrown there are 6 outcomes.

Let the probability of getting 4 on a throw which is termed as success is denoted by p.

⇒ p = 1/6

Let the probability of failure is denoted by q = 1 - p.

⇒ q = 1 -  1/6 = 5/6

As we know that, variance of a binomial distribution = npq

\(\rm ⇒variance = 72 \times \frac{1}{6} \times \frac{5}{6}=10\)

Hence, the correct option is 3.

Concept:

Binomial Distribution: 

If ‘n’ and ‘p’ are the parameters, then ‘n’ denotes the total number of times the experiment is conducted and ‘p’ denotes the probability of the happening of the event.

The probability of getting exactly ‘k’ successes in ‘n’ independent trials for a Random Variable X is expressed as P(X = k) and is given by the formula:

P(X = k) = nCk pk (1 - p)n - k

Calculation:

The probability of getting a head in a single toss is p = \(\frac12\).

∴ The probability of getting 2 heads in 6 tosses, will be:

P(X = 2) = 6C2\(\left(\frac12\right)^2\)\(\left(1-\frac12\right)^{6-2}\)

= \(\frac{15}{64}\).

Concept:

Binomial Distribution: A Binomial Distribution is an experiment in which only one out of two outcomes is possible. For example: Head or Tail, Yes or No, 1 or 0, etc.

• It is denoted as B(n, p) where ‘n’ denotes the total number of times the experiment is conducted and ‘p’ denotes the probability of the happening of the event.
• The probability of getting exactly ‘k’ successes in ‘n’ independent trials for a Random Variable X is expressed as P(X = k) and is given by the formula: P(X = k) = ⁿC_k p^k (1 - p)^{n - k}.
• ⁿC_k is maximum when k = \(\rm \dfrac n 2\), if n is even.
• nC_k is maximum for both k = \(\rm \dfrac {n+1} 2\) and k = \(\rm \dfrac {n+3} 2\), if n is odd.

Calculation:

Variable X has a binomial distribution \(\rm B\left(6, \dfrac{1}{2}\right)\).

∴ n = 6 and p = \(\rm \dfrac {1} 2\).

And, 1 - p = \(\rm 1-\dfrac {1} 2=\dfrac {1} 2\) = p.

For a random variable X, the probability is given by P(X = k) = nCk pk (1 - p)n - k.

∴ P (X = k) = 6Ck pk p6 - k

= 6Ck p6

= \(\rm ^6C_k\left(\dfrac {1}{2}\right)^6\)

Now, the most likely outcome will be when P (X = k) will be maximum for k = 0, 1, 2, 3, 4, 5 or 6.

⇒ \(\rm ^6C_k\left(\dfrac {1}{2}\right)^6\) is maximum.

⇒ 6Ck is maximum.

⇒ k = \(\rm \dfrac {6}{2}\) = 3.

∴ The most likely outcome is when the value of X = k = 3.

Explanation:

P(2 boys and 2 girls) \(= {\rm{\;P}}\left( {{\rm{X\;}} = {\rm{\;}}2} \right) = 4{C_2}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^{4 - 2}} = 6 \times {\left( {\frac{1}{2}} \right)^4}\)

∴ P(2 boys and 2 girls) \(= {\rm{\;P}}\left( {{\rm{X\;}} = {\rm{\;}}2} \right) = \frac{3}{8}\)

Now,

Number of families having 2 boys and 2 girls = N ⋅ P(X = 2)

(where N is the total number of families considered)

∴ Number of families having 2 boys and 2 girls \( = 800 \times \frac{3}{8}\)

np = 4 and npq = 3

thus q = (3/4) and p = (1-q) = (1/4)

mode is an integer such that np+p > x > np-q

⇒ 4 +(1/4) > x >4-(3/4)

⇒ (13/4)

3.25

⇒ x=4