Question

A coin will be tossed 7 times. Find the probability that there will be exactly 2 heads among the first 4 tosses, and exactly 2 heads among the last 3 tosses. (Include 2 digits after the decimal point.)

Answer 1

Let X denote the number of heads on the first 4 tosses of the coin and Y denote the number of heads on the last 3 tosses of the coin.

We know that each toss of a coin is independent from which we can conclude that the first 4 tosses and the last 3 tosses (of the 7 total tosses) are independent. Thus, we can conclude that X, which depends on the first 4 tosses, and Y, which depends on the last 3 tosses, are independent.

Now, the probability of getting heads on a single toss of a coin is 0.5.

Consider X. There are a fixed (four) number of tosses (fixed number of trials), each toss has two outcomes (head and tail) and on each toss the probability of getting a head is 0.5 independent of other tosses, thus we can conclude that:

X ~ Bin(n = 4, p = 0.5) and the probability mass function of X is given by:

Consider Y. There are a fixed (three) number of tosses (fixed number of trials), each toss has two outcomes (head and tail) and on each toss the probability of getting a head is 0.5 independent of other tosses, thus we can conclude that:

Y ~ Bin(N = 3, q = 0.5) and the probability mass function of Y is given by:

Now, the probability that there will be exactly 2 heads among the first 4 tosses, and exactly 2 heads among the last 3 tosses is given by:

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