Question

From a group of 10 children, a team of 5 (the red team) is randonnly chosen. The remaining 5 children

Answer 1

Using the concept of probability and the combination formula, it is found that there is a 0.4444 = 44.44% probability  they get to be on the same team.

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• A probability is the number of desired outcomes divided by the number of total outcomes.
• The order in which the children are chosen is not important, which means that the combination formula is used to find the number of outcomes.

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Combination formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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Finding the number of desired outcomes:

• Andy and Ollie on the same team, plus 3 children from a set of 8.
• Can be on either team, blue or red, so multiplied by 2.

$D = 2C_{8,3} = 2\frac{8!}{3!5!} = 112$

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Finding the number of total outcomes:

• 5 children from a set of 10, thus:

$T = C_{10,5} = \frac{10!}{5!5!} = 252$

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The probability is:

$p = \frac{D}{T} = \frac{112}{252} = 0.4444$

0.4444 = 44.44% probability  they get to be on the same team.

A similar problem is given at brainly.com/question/22931444