Answer:
The required probability is 0.127.
Step-by-step explanation:
The binomial model is given by:
[tex]P(X = x)=nC_{x}p^{n-x} q^{x}[/tex]
where p is the probability of a success and q is the probability of a failure.
q = 1 - p
= 1 - 0.533
= 0.467
Substitute p = 0.533, q = 0.467 and n = 5, x = 4, we get,
[tex]P(X=4)=5C_{4}(0.533)^{1}(0.467)^{4}[/tex]
[tex]=5(0.533)(0.467)^{4}[/tex]
≅ 0.127
Answer:
0.128
Step-by-step explanation:
We know the probability for any event X is given by,
[tex]P(X=x)=\binom{n}{x}\times p^{n-x}\times q^{x}[/tex],
where p is the probability of success and q is the probability of failure.
Here, we are given that p = 0.533.
Since, we have that q = 1 - p
i.e. q = 1 - 0.533
i.e. q = 0.467
It is required to find the probability of 4 wins in the next 5 games i.e. P(X=4) when n = 5.
Substituting the values in the above formula, we get,
[tex]P(X=4)=\binom{5}{4}\times 0.533^{5-4}\times 0.467^{4}[/tex]
i.e. [tex]P(X=4)=5 \times 0.533 \times 0.048[/tex]
i.e. [tex]P(X=4)=5 \times 0.533 \times 0.048[/tex]
i.e. i.e. [tex]P(X=4)=0.128[/tex]
Hence, the probability of 4 wins in the next 5 games is 0.128.
