A certain basketball player scores 60% of the free-throw shots she attempts. During a particular game, she gets six free throws. (a) What assumption(s) must be made in order to regard this as a sequence of Bernoulli trials? (b) Based on such assumption(s), find, correct to three decimal places, the probability that: (i) she scores on exactly four of the six shots (ii) she scores for the second time on the fifth shot.

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A certain basketball player scores 60% of the free-throw shots she attempts - Leaving Cert Mathematics - Question 4 - 2012

Question 4

A certain basketball player scores 60% of the free-throw shots she attempts. During a particular game, she gets six free throws. (a) What assumption(s) must be made... show full transcript

Worked Solution & Example Answer:A certain basketball player scores 60% of the free-throw shots she attempts - Leaving Cert Mathematics - Question 4 - 2012

Step 1

What assumption(s) must be made in order to regard this as a sequence of Bernoulli trials?

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Answer

To regard the situation as a sequence of Bernoulli trials, the following assumptions must be made:

Independence of Trials: Each free throw attempt must be independent of the others, meaning the outcome of one throw does not influence the outcome of another.
Constant Probability of Success: The probability of scoring (success) remains the same for each throw. In this case, the player scores 60% of the time, so the probability of success for each shot is 0.6.
Two Possible Outcomes: Each shot has only two possible outcomes: success (she scores) or failure (she misses).
Finite Number of Throws: The number of free throws taken must be finite, which in this scenario is six.

Step 2

she scores on exactly four of the six shots

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Answer

To find the probability that she scores on exactly four of the six shots, we can use the binomial probability formula:

$P(X = k) = {n race k} p^k (1-p)^{n-k}$

Where:

$n = 6$ (total throws)
$k = 4$ (successful throws)
$p = 0.6$ (probability of success)

Calculating: $P(X = 4) = {6 race 4} (0.6)^4 (0.4)^{2}$

The binomial coefficient ${6 race 4} = 15$ , so we have:

$P(X = 4) = 15 (0.6)^4 (0.4)^{2}$ $P(X = 4) = 15 (0.1296) (0.16)$ $P(X = 4) = 15 (0.020736) = 0.31104$

Thus, rounding to three decimal places, the probability is 0.311.

Step 3

she scores for the second time on the fifth shot

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Answer

To find the probability that she scores for the second time on the fifth shot, we first note that she must score exactly once in the first four shots and then score on the fifth shot. Thus, we get:

One success among the first four throws: This follows a binomial distribution.
The probability of scoring on the fifth shot is simply the probability of success, which is 0.6.

The probability of one score in the first four throws is given by: $P(X = 1) = {4 race 1} (0.6)^{1} (0.4)^{3}$

Calculating: $P(X = 1) = 4 imes (0.6) imes (0.4)^{3}$ $= 4 imes 0.6 imes 0.064 = 0.1536$

Thus, the overall probability is: $P = P(X=1 ext{ in first 4}) imes P( ext{success on fifth})$ $P = 0.1536 imes 0.6 = 0.09216$

Rounding to three decimal places, the probability is 0.092.

A certain basketball player scores 60% of the free-throw shots she attempts - Leaving Cert Mathematics - Question 4 - 2012

Worked Solution & Example Answer:A certain basketball player scores 60% of the free-throw shots she attempts - Leaving Cert Mathematics - Question 4 - 2012

What assumption(s) must be made in order to regard this as a sequence of Bernoulli trials?

she scores on exactly four of the six shots

she scores for the second time on the fifth shot

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