SL 4.5—Trial and outcome

The coin has 2 outcomes and the die has 6 outcomes.

For combined independent experiments, multiply the number of outcomes: $n(U)=2\times 6$.

$n(U)=12$.

The sample space contains 12 outcomes such as (H,1), (H,2), ..., (T,6).

This would be $2+6$, adding instead of multiplying the outcomes.

This is just the die's sample space, ignoring the coin entirely.

This is $6\times 6$, treating both experiments as dice instead of one coin and one die.

The correct combined sample space size is 12, option B.

$n(U)=6+4+10=20$.

$P(\text{white})=\frac{10}{20}$.

$P(\text{not white})=1-\frac{10}{20}=\frac{10}{20}$.

The probability of not drawing white equals the probability of drawing orange or purple, which is $6+4=10$ out of 20.

This incorrectly uses the number of white marbles as the denominator instead of the total.

This only counts orange marbles, forgetting to add purple.

This only counts purple marbles, forgetting to add orange.

Combining orange and purple gives $10/20$, option A.