Winning Secrets in Coin Toss

Solution & Explanation

To tackle the problem effectively, let's break down the scenario and apply probability theory to understand the outcome.

Problem Breakdown

1. Initial Setup:

   • 100 students participate in a coin-tossing challenge.
   • Each student flips a coin.
   • If a student's coin lands on heads, they win.
   • If it lands on tails, they flip again.
   • On the second flip, if the coin lands on heads, they falsely claim victory.
   • If it lands on tails again, they admit defeat.

2. Outcome:

   • 30 students claim to have won by the end.
   • We need to determine how many students truly won.

Probability Analysis

To solve this, consider the probability of each possible outcome:

• True Win (H): Probability of winning on the first flip is $P(H) = \frac{1}{2}$.
• False Win (T followed by H, or TH): Probability of losing the first flip and winning on the second flip is $P(TH) = \frac{1}{4}$.
• True Loss (TT): Probability of losing both flips is $P(TT) = \frac{1}{4}$.

Calculating Proportions

Given these probabilities, the total probability of declaring a win (either true or false) is:

$P(\text{Declaring a Win}) = P(H) + P(TH) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$

Thus, the probability of a student truly winning (given that they declared a win) is:

$P(\text{True Win | Declaring a Win}) = \frac{P(H)}{P(\text{Declaring a Win})} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$

Expected Number of True Winners

If 30 students declared a win, then the expected number of students who truly won is:

$E[\text{True Winners}] = 30 \times \frac{2}{3} = 20$

Conclusion

Therefore, given the problem's conditions and the calculations above, we can reasonably expect that 20 students truly won the coin-tossing challenge. This solution leverages the understanding of probability distributions and conditional probability to arrive at the expected number of true winners.