Suppose that the New England Colonials baseball team is equally likely to win a

sanuluy 2021-10-02 Answered
Suppose that the New England Colonials baseball team is equally likely to win a game as not win it. If 4 Colonials games are chosen at random, what is the probability that exactly 3 of those games are won by the Colonials?
Round your response to at least three decimal places. (If necessary, consult a list of formulas.)

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Expert Answer

au4gsf
Answered 2021-10-03 Author has 9833 answers
Step 1
Given Information :
Suppose that the New England Colonials baseball team is equally likely to win a game as not to win it. 4 Colonials games are chosen at random.
Let, p denotes probability of winning a game .
\(\displaystyle{p}={0.5}\) (equally likely)
n denotes sample size
\(\displaystyle{n}={4}\)
Define random variable X denotes number of games are won by colonials .
Since, events are independent . \(\displaystyle{X}\sim{B}{\left({n}={4},{p}={0.5}\right)}\)
Binomial Probability formula is given as:
\(\displaystyle{P}{\left({X}\right)}={\frac{{{n}!}}{{{X}!{\left({n}-{X}\right)}!}}}\cdot{p}^{{{X}}}\cdot{\left({1}-{p}\right)}^{{{n}-{X}}}\)
Step 2
Exactly 3 of those games are won by the Colonials i.e. \(\displaystyle{X}={3}\)
\(\displaystyle{P}{\left({X}={3}\right)}\) Probability of exactly 3 successes
If using a calculator, you can enter trials \(\displaystyle={4},{p}={0.5},{\quad\text{and}\quad}\ {X}={3}\) into a binomial probability formula.
\(\displaystyle{P}{\left({X}\right)}={\frac{{{n}!}}{{{X}!{\left({n}-{X}\right)}!}}}\cdot{p}^{{{X}}}\cdot{\left({1}-{p}\right)}^{{{n}-{X}}}\)
where n is the number of trials, p is the probability of success on a single trial, and X is the number of successes. Substituting in values for this problem, \(\displaystyle{n}={4},{p}={0.5},{\quad\text{and}\quad}\ {X}={3}\).
\(\displaystyle{P}{\left({3}\right)}={\frac{{{4}!}}{{{3}!{\left({4}-{3}\right)}!}}}\cdot{0.5}^{{{3}}}\cdot{\left({1}-{0.5}\right)}^{{{4}-{3}}}\)
Evaluating the expression, we have
\(\displaystyle{P}{\left({X}={3}\right)}={0.25}\)
Probability that exactly 3 of those games are won by the Colonials \(\displaystyle={0.25}\).
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