 # 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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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 $$\displaystyle={0.25}$$.