Imagine that the New England Colonials baseball team has an equal chance of winning and losing a game. What is the likelihood that, out of four games chosen at random, exactly three will be won by the Colonials?Round your response to at least three decimal places. (If necessary, refer 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 .   p=0.5 (equally likely)  n denotes sample size   n=4   Define random variable X denotes number of games are won by colonials .   Since, events are independent . X∼B(n=4,p=0.5)   Binomial Probability formula is given as:  P(X)=n!X!(n−X)!⋅pX⋅(1−p)n−X  Step 2  Exactly 3 of those games are won by the Colonials i.e. X=3   P(X=3) Probability of exactly 3 successes   If using a calculator, you can enter trials =4,p=0.5,and X=3 into a binomial probability formula.  P(X)=n!X!(n−X)!⋅pX⋅(1−p)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, n=4,p=0.5,and X=3.  P(3)=4!3!(4−3)!⋅0.53⋅(1−0.5)4−3  Evaluating the expression, we have  P(X=3)=0.25   Probability that exactly 3 of those games are won by the Colonials =0.25.

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

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