Step-by-step explanation:
<em>"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."</em>
<em />
<em>"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."</em>
<em>"(i) Find the total number of possible ways that these six pairs can be formed."</em>
The order doesn't matter (AB is the same as BA), so use combinations.
For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.
For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.
So on and so forth. The total number of combinations is:
₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂
= 66 × 45 × 28 × 15 × 6 × 1
= 7,484,400
<em>"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."</em>
We need to find the number of ways that 6 boys can be grouped into 3 pairs. Using the same logic as before:
₆C₂ × ₄C₂ × ₂C₂
= 15 × 6 × 1
= 90
There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs. So the probability is:
90 × 90 / 7,484,400
= 1 / 924
≈ 0.0011
<em>"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.</em>
<em>(i) How many different schedules are possible?"</em>
There are ₁₂C₆ ways that 6 players can be selected from 12. From there, each possible schedule has a different order of players, so we need to use permutations.
There are 6 options for the first match. After that, there are 5 options for the second match. Then 4 options for the third match. So on and so forth. So the number of permutations is 6!.
The total number of possible schedules is:
₁₂C₆ × 6!
= 924 × 720
= 665,280
<em>"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."</em>
If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.
If 0 girls are selected, the number of combinations is:
₆C₆ × ₆C₀ = 1 × 1 = 1
If 1 girl is selected, the number of combinations is:
₆C₅ × ₆C₁ = 6 × 6 = 36
If 2 girls are selected, the number of combinations is:
₆C₄ × ₆C₂ = 15 × 15 = 225
If 3 girls are selected, the number of combinations is:
₆C₃ × ₆C₃ = 20 × 20 = 400
If 4 girls are selected, the number of combinations is:
₆C₂ × ₆C₄ = 15 × 15 = 225
The probability that there are more boys than girls is:
(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)
= 262 / 887
≈ 0.2954
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