Five discs numbered 1, 3, 4, 6 and 7 are placed in a bag. Two disc are taken out of the bag at random without replacement.

1. Find the probabilities for the first draw

There are $3$ odd discs: $1,3,7$, and $2$ even discs: $4,6$.

Total number of discs $=5$.

So
$P(\text{Odd on first draw})=\frac35,$
$P(\text{Even on first draw})=\frac25.$

2. Find the conditional probabilities for the second draw without replacement

If the first disc is odd, there are $4$ discs left: $2$ odd and $2$ even.

So
$P(\text{Odd second}\mid \text{Odd first})=\frac24=\frac12,$
$P(\text{Even second}\mid \text{Odd first})=\frac24=\frac12.$

If the first disc is even, there are $4$ discs left: $3$ odd and $1$ even.

So
$P(\text{Odd second}\mid \text{Even first})=\frac34,$
$P(\text{Even second}\mid \text{Even first})=\frac14.$

1. Add the probabilities of the two favourable outcomes

One disc is odd and the other is even means either:

$\text{Odd then Even}$ or $\text{Even then Odd}$.

$P(\text{one odd and one even})=\frac35\times\frac12+\frac25\times\frac34$

$=\frac3{10}+\frac3{10}=\frac6{10}=\frac35$

1. Find and compare the probability that both numbers drawn are less than 4

The numbers less than $4$ are $1$ and $3$.

$P(\text{both less than }4)=\frac{2}{5}\times\frac{1}{4}=\frac{1}{10}=10\%.$

So it is exactly $10\%$, not more than $10\%$. Peter is wrong.

1. List all possible pairs and identify those with a prime sum

There are $\binom{5}{2}=10$ possible pairs.

The pairs with a prime sum are
$(1,4),\ (1,6),\ (3,4),\ (4,7),\ (6,7),\ (3,6).$
Their sums are respectively $5,7,7,11,13,9$.

Since $9$ is not prime, the favourable pairs are
$(1,4),\ (1,6),\ (3,4),\ (4,7),\ (6,7).$
So there are $6$ favourable ordered outcomes, or equivalently $5$ favourable unordered pairs.

2. Calculate the probability of a prime sum

Using ordered outcomes, there are $5\times4=20$ possible outcomes.

The $5$ favourable unordered pairs give $10$ favourable ordered outcomes.

$P(\text{prime sum})=\frac{10}{20}=\frac12$