1. **State the problem:** Anna has 2 coins totaling 30p, and Tom has 4 coins totaling 30p. We want to find who is more likely to pick a 10p coin from their respective bags. 2. **Analyze Anna's bag:** Let the number of 10p coins Anna has be $x$. Since she has 2 coins total, the other coin's value is $30 - 10x$ pence. Possible values for $x$ are 0, 1, or 2. - If $x=0$, total value would be less than 30p (impossible). - If $x=1$, the other coin is $30 - 10 = 20$p (a 20p coin). - If $x=2$, total value is $2 \times 10 = 20$p, which is less than 30p (impossible). So Anna must have exactly one 10p coin and one 20p coin. Probability Anna picks a 10p coin is: $$P_A = \frac{1}{2} = 0.5$$ 3. **Analyze Tom's bag:** Tom has 4 coins totaling 30p. Let the number of 10p coins be $y$. The total value of the 4 coins is: $$10y + \text{value of other coins} = 30$$ Since coins are standard UK coins, possible coins are 1p, 2p, 5p, 10p, 20p, 50p, etc. But since total is 30p with 4 coins, and 10p coins are involved, let's consider possible $y$ values. - If $y=0$, other 4 coins total 30p. - If $y=1$, other 3 coins total 20p. - If $y=2$, other 2 coins total 10p. - If $y=3$, other 1 coin total 0p (impossible). - If $y=4$, total 40p (too high). Try $y=1$: Other 3 coins total 20p, possible with 5p, 5p, 10p or 10p, 5p, 5p, but that would increase 10p coins. Try $y=2$: Other 2 coins total 10p, possible with 5p and 5p. So Tom likely has 2 coins of 10p and 2 coins of 5p. Probability Tom picks a 10p coin is: $$P_T = \frac{2}{4} = 0.5$$ 4. **Compare probabilities:** $$P_A = 0.5, \quad P_T = 0.5$$ Both Anna and Tom are equally likely to pick a 10p coin. **Final answer:** Both equally likely.