Question

Find the binomial distribution for flipping a coin 3 times, where “heads” is a success. P(k successes) = nCk pk(1 – p)n – k P(0 successes) = 3C0(0.5)0(0.5)3 = P(1 success) = P(2 successes) = P(3 successes) =

Answer 1

Answer:

$P(0\ successes)=0.125\\\\P(1\ successes)=0.375\\\\P(2\ successes)=0.375\\\\P(3\ successes)=0.125$

Step-by-step explanation:

We know that the binomial distribution for k successes out of n experiments is given as:

$P(k\ successes) = n_C_k\cdot p^k\cdot (1-p)^{n-k}$

Here we have k= number of heads.

p=probability of getting a head=1/2=0.5

and n=number of trials=3.

Hence,

$P(0 successes)=3_c_0\cdot(0.5)^{0}\cdot (1-0.5)^{3-0}\\\\P(0\ successes)=3_c_0\cdot 1\cdot (0.5)^{3}\\\\P(0\ successes)=0.125$

Similarly,

$P(1 successes)=3_c_1\cdot(0.5)^{1}\cdot (1-0.5)^{3-1}\\\\P(1\ successes)=3\cdot 0.5\cdot (0.5)^{2}\\\\P(0\ successes)=0.375$

Similarly,

$P(2 successes)=3_c_2\cdot(0.5)^{2}\cdot (1-0.5)^{3-2}\\\\P(0\ successes)=3\cdot (0.5)^{2}\cdot (0.5)^{1}\\\\P(0\ successes)=0.375$

Similarly,

$P(3 successes)=3_c_3\cdot(0.5)^{3}\cdot (1-0.5)^{3-3}\\\\P(3\ successes)=1\cdot (0.5)^3\cdot (0.5)^{0}\\\\P(3\ successes)=0.125$

Hence,

$P(0\ successes)=0.125\\\\P(1\ successes)=0.375\\\\P(2\ successes)=0.375\\\\P(3\ successes)=0.125$

Answer 2

Probability of an events is its chance of occurrence .The probabilities of flipping heads are

• P(k successes) = $P(X =k) = \: ^nC_kp^k(1-p)^{n-k}$
• P(0 success) = $P(X =0) = 0.125$
• P(1 success) =  $P(X = 1) = 0.375$
• P(2 success) =  $P(X =2) = 0.375$
• P(3 success) =  $P(X =3) = 0.125$

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be k successes is given by

$P(X =k) = \: ^nC_kp^k(1-p)^{n-k}$

For the given case, the number of Bernoulli trials are 3.

Each coin have half probability for head or tail outcome(as coins are fair(assumed) and only two outcomes, so they take half of the total probability which comes as 0.5)

Thus, p = 1 - p = 0.5

Thus, if X is tracking the number of "heads" in this case, we get:

$X \sim B(n,p)\\X \sim B(3, 0.5)$

Using the above probability function, we get the probabilities as:

• Case 1 : k = 0

$P(X =0) = \: ^3C_0p^0(1-p)^{3-0} = ( 1 - p)^ 3 = (1-p)^3 = (0.5)^3 = 0.125$

• Case 2: k = 1

$P(X =1) = \: ^3C_1p^1(1-p)^{3-1} = 3 \times (0.5)( 0.5)^2 = 3 \times(0.5)^3 = 0.375$

• Case 3: k = 2

$P(X =2) = \: ^3C_2p^1(1-p)^{3-2} = 3 \times (0.5)^2( 0.5) = 3 \times(0.5)^3 = 0.375$

• Case 4: k = 3

$P(X =3) = \: ^3C_3p^1(1-p)^{3-3} = 1 \times (0.5)^3( 0.5)^0 = (0.5)^3 = 0.125$

Thus,

The answers are

• P(k successes) = $P(X =k) = \: ^nC_kp^k(1-p)^{n-k}$
• P(0 success) = $P(X =0) = 0.125$
• P(1 success) =  $P(X = 1) = 0.375$
• P(2 success) =  $P(X =2) = 0.375$
• P(3 success) =  $P(X =3) = 0.125$

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