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

0 successes= 0.125
1 success=  0.375
2 successes= 0.375
3 successes= 0.125

Answer 2

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,

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