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Vinvika [58]
3 years ago
9

Four couples at a dinner party play a board game after the meal. They decide to play as teams of two and to select the teams ran

domly. All eight people write their names on slips of paper. The slips are thoroughly mixed, then drawn two at a time. How likely is it that every person will be teamed with someone other than the person he or she came to the party with?
Mathematics
1 answer:
Llana [10]3 years ago
4 0

Answer:

0.29

Step-by-step explanation:

We are given that there are 4 couples i.e. 8 people

They decide to play as teams of two and to select the teams randomly

Now we are asked How likely is it that every person will be teamed with someone other than the person he or she came to the party with

For people 1 , there are 6 options for pairing Since he or she cannot pair with his /her own partner

So, Choices For people 1 will be 6 person

So, Probability that person 1 will be teamed with someone other than the person he or she came to the party with = \frac{\text{No. of choices for Person 1}}{\text{Total no. of choices available}}=\frac{6}{7}

So,  Probability that every person (=8 person) will be teamed with someone other than the person he or she came to the party with =(\frac{6}{7})^8=0.29

Hence Probability that every person will be teamed with someone other than the person he or she came to the party with is 0.29

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Answer:

Georgina should add 3 liters of 49% citric acid solution.

Step-by-step explanation:

let x be the liters of 49% solution that Georgina should add. Then after she has added 0.5 liters of water, the total liters of solution she has will be

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snow_tiger [21]

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Step-by-step explanation:

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