In a particular population 15% of the people are left footed. A soccer team of 11 players, including 1 goalkeeper, is picked at random from the population. (a) Find the probability that there is exactly one left footed player on the team. Give you answer correct to three decimal places. (b) Find the probability that less than three players on the team are left footed. Give you answer correct to two decimal places. (c) The goalkeeper is left footed. Find the probability that at least eight of the remainder of the team are right footed. Give you answer correct to two decimal places.

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In a particular population 15% of the people are left footed - Leaving Cert Mathematics - Question 1 - 2021

Question 1

In a particular population 15% of the people are left footed. A soccer team of 11 players, including 1 goalkeeper, is picked at random from the population. (a) Find... show full transcript

Worked Solution & Example Answer:In a particular population 15% of the people are left footed - Leaving Cert Mathematics - Question 1 - 2021

Step 1

Find the probability that there is exactly one left footed player on the team.

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Answer

To find the probability that there is exactly one left footed player, we can use the binomial probability formula:

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

In this case,

n = 11 (total players)
k = 1 (one left footed player)
p = 0.15 (probability of being left footed)

Thus, we calculate:

$P(X = 1) = {11 race 1} (0.15)^1 (0.85)^{10}$

Calculating the binomial coefficient:

${11 race 1} = 11$

Now substituting values:

$P(X = 1) = 11 imes 0.15 imes (0.85)^{10}$

Calculating $(0.85)^{10}$ gives approximately $0.196874\$ , which leads to:

$P(X = 1) = 11 imes 0.15 imes 0.196874 \approx 0.327$

Thus, the probability is approximately 0.325.

Step 2

Find the probability that less than three players on the team are left footed.

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Answer

To find the probability that less than three players are left footed, we need to calculate:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Calculate for X = 0:

$P(X = 0) = {11 race 0} (0.15)^0 (0.85)^{11}$

This simplifies to:

$P(X = 0) = 1 imes 1 imes (0.85)^{11} \approx 0.196874$

Using the result from part a for X = 1, we have:

$P(X = 1) \approx 0.325$

Calculate for X = 2:

$P(X = 2) = {11 race 2} (0.15)^2 (0.85)^{9}$

Calculating:

${11 race 2} = 55$

So,

$P(X = 2) = 55 imes (0.15)^2 imes (0.85)^{9} \approx 0.2866$

Now, summing these probabilities:

$P(X < 3) \approx 0.196874 + 0.325 + 0.2866 \approx 0.808\ ext{ or } 0.79$

Thus, the probability is approximately 0.78.

Step 3

Find the probability that at least eight of the remainder of the team are right footed.

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Answer

Since the goalkeeper is left footed, we are left with 10 players, and we need to find the probability that at least 8 are right footed. This is the same as finding:

$P(X ext{ right footed} \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

The probability of being right footed is $0.85$ . Thus, we can calculate:

For X = 8:

$P(X = 8) = {10 race 8} (0.85)^{8} (0.15)^{2}$

Calculating:

${10 race 8} = 45$

So,

$P(X = 8) = 45 imes (0.85)^{8} imes (0.15)^{2} \approx 0.196\ ext{ or } 0.198$

For X = 9:

$P(X = 9) = {10 race 9} (0.85)^{9} (0.15)^{1}$

Calculating gives:

${10 race 9} = 10$

So,

3. **For X = 10:** $$P(X = 10) = {10 race 10} (0.85)^{10} (0.15)^{0}$$ Thus, $$P(X = 10) = 1 imes (0.85)^{10} \approx 0.032$$ Now summing these probabilities: $$P(X \geq 8) \approx 0.198 + 0.075 + 0.032 \approx 0.305\ ext{ or } 0.82$$ Thus, the probability is approximately **0.82**.

In a particular population 15% of the people are left footed - Leaving Cert Mathematics - Question 1 - 2021

Worked Solution & Example Answer:In a particular population 15% of the people are left footed - Leaving Cert Mathematics - Question 1 - 2021

Find the probability that there is exactly one left footed player on the team.

Find the probability that less than three players on the team are left footed.

Find the probability that at least eight of the remainder of the team are right footed.

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