5. A couple decides to have 3 children. If none of the 3 is a girl, they will try again; and if they still don't get a girl, they will try once more. If the random variable $$ X $$ denotes the number of children the couple will have following this scheme, then what is the expected value of $$ X $$ ?

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To find the expected value of the random variable $$ X $$, which represents the number of children the couple will have following the given scheme, we need to consider the probabilities associated with each possible outcome.

Given that the couple will try again if none of the 3 children is a girl, and they will try once more if they still don't get a girl, we can break down the possible outcomes as follows:

1. The couple has 3 children, all of whom are boys.
2. The couple has 4 children, with 3 boys and 1 girl.
3. The couple has 5 children, with 4 boys and 1 girl.

Let's calculate the probabilities for each of these outcomes:

1. The probability of having 3 boys in the first 3 children is $$\left(\frac{1}{2}\right)^3$$.
2. The probability of having 3 boys in the first 3 children and then a girl in the 4th child is $$\left(\frac{1}{2}\right)^4$$.
3. The probability of having 3 boys in the first 3 children, a girl in the 4th child, and then a boy in the 5th child is $$\left(\frac{1}{2}\right)^5$$.

Now, we can calculate the expected value of $$ X $$ by multiplying each outcome by its probability and summing them up:

$$ E(X) = 3 \times \left(\frac{1}{2}\right)^3 + 4 \times \left(\frac{1}{2}\right)^4 + 5 \times \left(\frac{1}{2}\right)^5 $$

Let's calculate this expression.
Calculate the value by following steps:
- step0: Calculate:
  $$3\left(\frac{1}{2}\right)^{3}+4\left(\frac{1}{2}\right)^{4}+5\left(\frac{1}{2}\right)^{5}$$
- step1: Multiply the terms:
  $$\frac{3}{8}+4\left(\frac{1}{2}\right)^{4}+5\left(\frac{1}{2}\right)^{5}$$
- step2: Multiply the terms:
  $$\frac{3}{8}+\frac{1}{2^{2}}+5\left(\frac{1}{2}\right)^{5}$$
- step3: Multiply the terms:
  $$\frac{3}{8}+\frac{1}{2^{2}}+\frac{5}{32}$$
- step4: Evaluate the power:
  $$\frac{3}{8}+\frac{1}{4}+\frac{5}{32}$$
- step5: Reduce fractions to a common denominator:
  $$\frac{3\times 4}{8\times 4}+\frac{2\times 4}{4\times 2\times 4}+\frac{5}{32}$$
- step6: Multiply the numbers:
  $$\frac{3\times 4}{32}+\frac{2\times 4}{4\times 2\times 4}+\frac{5}{32}$$
- step7: Multiply the terms:
  $$\frac{3\times 4}{32}+\frac{2\times 4}{32}+\frac{5}{32}$$
- step8: Transform the expression:
  $$\frac{3\times 4+2\times 4+5}{32}$$
- step9: Multiply the numbers:
  $$\frac{12+2\times 4+5}{32}$$
- step10: Multiply the numbers:
  $$\frac{12+8+5}{32}$$
- step11: Add the numbers:
  $$\frac{25}{32}$$
The expected value of the random variable $$ X $$ is $$ \frac{25}{32} $$ or approximately 0.78125.
The expected value of $$ X $$ is $$ \frac{25}{32} $$ or approximately 0.78125.

5. A couple decides to have 3 children. If none of the 3 is a girl, they will try again; and if they still don't get a girl, they will try once more. If the random variable \( X \) denotes the number of children the couple will have following this scheme, then what is the expected value of \( X \) ?

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