Question

If a fair coin is tossed 4 times, there are 16 possible sequences of heads (H) and tails (T). Supposethe random variable X represents the number of heads in a sequence. Construct the probabilitydistribution for X.

Answer 1

Answer:

The probability distribution of X is:

$P(x=0)=0.0625\\\\P(x=1)=0.25\\\\P(x=2)=0.375\\\\P(x=3)=0.25\\\\P(x=4)=0.0625$

Step-by-step explanation:

"Tossing a coin and getting H" can be modelled as a Bernoulli random variable. The variable X is a sum of 4 of this Bernoulli variables, we can model X as a binomial random variable, with p=0.5 and n=4.

The values X can take are: 0, 1, 2, 3 and 4.

The values of this probability are calculated as:

$P(x=0) = \binom{4}{0} p^{0}q^{4}=1*1*0.0625=0.0625\\\\P(x=1) = \binom{4}{1} p^{1}q^{3}=4*0.5*0.125=0.25\\\\P(x=2) = \binom{4}{2} p^{2}q^{2}=6*0.25*0.25=0.375\\\\P(x=3) = \binom{4}{3} p^{3}q^{1}=4*0.125*0.5=0.25\\\\P(x=4) = \binom{4}{4} p^{4}q^{0}=1*0.0625*1=0.0625$

Where

p: is the probability of tossing a coin and getting a Heads

q: is the probability of tossing a coin and not getting a Heads.

The binomial number is a way to calculate the possible combinations of getting a certain amount of heads.

For example, in 4 tosses, there are 6 ways or combinations of getting 2 heads and 2 tails:

$\binom{4}{2}=\frac{4!}{2!(4-2)!}=\frac{24}{2*2}=6$