Suppose $X$ and $Y$ are jointly normal, i.e. no independence is needed. Define $Z = X - Y$. It is well known that $Z$ is Gaussian, and thus is determined by its mean $\mu$ and its variance $\sigma^2$.
$$
\mu = \mathbb{E}(Z) = \mathbb{E}(X) - \mathbb{E}(Y) = \mu_1 - \mu_2
$$
$$
\sigma^2 = \mathbb{Var}(Z) = \mathbb{Var}(X) + \mathbb{Var}(Y) - 2 \mathbb{Cov}(X,Y) = \sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2
$$
where $\rho$ is the correlation coefficient. Now:
$$
\mathbb{P}(X>Y) = \mathbb{P}(Z>0) = 1- \Phi\left(-\frac{\mu}{ \sigma}\right) = \Phi\left(\frac{\mu}{ \sigma}\right) = \frac{1}{2} \operatorname{erfc}\left(-\frac{\mu}{\sqrt{2}\sigma}\right)
$$
The elegant answer by @Graham Kemp would be correct if the question were whether the
number in the of reds in the sample of 50 is less than the number
in the sample of 100. But the question was about proportions.
A simulation of a million repetitions of this experiment confirms
that the probability of the former is about 0.539 (last place in
some doubt). For proportions the number is much different: about 0.023.
This makes sense because there are (said to be) proportionately more reds in
child's population.
My guess is you are expected to use normal approximations for these
proportions.
For the child mix the number is $X \sim Bin(50,.3),$
so $E(X) = 50(.3) = 15$ and $V(X) = 15(.7) = 10.5$
For proportions $E(\hat p_c) = (X/n) = .3$ and $V(\hat p_c) = (.3)(.7)/50 = 0.0042.$
For proportions in the adult mix
$E(\hat p_a) = .15$ and $V(\hat p_a) = (.15)(.85)/100 = 0.001275.$
You are interested in the difference in proportions, which can be found
by subtracting the means and (by independence) adding the variances.
$E(\hat p_c - \hat p_a) = .3-.15 = .15$ and
$V(\hat p_c - \hat p_a) = .0042 + .001275 = 0.005475.$
The two binomial proportions are roughly normally distributed, and so
their difference is also. The question is whether the difference
in sample proportions is less than 0.
With software this can be computed directly as 0.0213.
On an exam you would have to standardize (subtracting the mean from 0 and dividing
the difference by the standard deviation) and look up the result in normal
tables. Some accuracy is lost in the necessary rounding. I got 0.0212.
The sample sizes are a little small for normal approximations to be
accurate, and in any case they can't be trusted beyond the second
decimal place. But the agreement with simulation results (which
suffer only from simulation approximation) is encouraging. In any
case, about the only computational option for an in-class exam
without computers is to use normal tables.
The bottom line is that it is extremely unlikely
the proportion of reds drawn from the child mix would be smaller
than the proportion drawn from the adult mix.
The R computer code for the simulation is as follows:
> m = 10^6; x = rbinom(m, 50, .3)/50; y = rbinom(m, 100, .15)/100
> mean(x <= y)
[1] 0.023093
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