In basketball, players often have to take free throws - Leaving Cert Mathematics - Question 8 - 2015

The probability that Michael is unsuccessful in his first two free throws and successful in the third can be calculated as:

$P(U, U, S) = 0.3 imes 0.4 imes 0.6 = 0.072$

The ways for Michael to be successful on his third throw:

1. Successful on first two throws: S, S, S
2. Successful on first, unsuccessful on second: S, U, S
3. Unsuccessful on first two, but successful on third: U, U, S

Calculating the probabilities for each path:

1. P(S, S, S) = 0.448
2. P(S, U, S) = 0.7 * 0.2 * 0.6 = 0.084
3. P(U, U, S) = 0.3 * 0.4 * 0.6 = 0.072

So, the total probability that he is successful on his third free throw:

$P = 0.448 + 0.072 + 0.084 = 0.604$

From the provided definitions, we can align as follows:

$p_{n+1} = P(S, S) + P(U, S) = p_n imes 0.8 + (1 - p_n) imes 0.6$ When simplified: $= p_n imes 0.8 + 0.6 - 0.6 imes p_n = 0.6 + 0.2p_n$

Assuming for large n, that $p_{n+1} ightarrow p$ and $p_n ightarrow p$, we can equate:

$p = 0.6 + 0.2p$ Solving yields:

Given that $a_n = p - p_n$ and $p_n$ = p thus:

rac{a_{n+1}}{a_n} = rac{p - p_{n+1}}{p - p_n} = rac{0.75 - (0.6 + 0.2p_n)}{0.75 - p_n} = rac{0.15 - 0.2p_n}{0.75 - p_n} By substitution, as $n$ grows, this ratio approaches $1/5$. Thus confirmed the geometric sequence.

Converting the sequence: $a_1 = p - p_1 = 0.75 - 0.75 = 0.05$ Then apply for limit convergence: $0.05(1/5)^{n-1} < 0.00001$ Taking logarithms: $(n-1) ln(1/5) < ln(0.00001) - ln(0.05)$ Results:

Given his previous success rate, the best estimate of success would be:

• 0.75 or p

This is because events are not independent; the outcome of one throw influences the probability of success in the subsequent throws.