Answer:
2 a. 210 ways
b. 50 ways
c. 28 ways
d. 0.205
4 a. 1716 ways
Step-by-step explanation:
2. a. The number of ways, [tex]n_w[/tex] of selecting without restriction is given as follows;
[tex]\dbinom{10}{4}[/tex] = 10!/(4!(10 - 4)!) = 210
b. The number of ways of selecting 3 boys from 5 is 5!/(3!(5 - 3)!) = 10 ways
The number of ways of selecting the remaining 1 girl from 5 = 5!/(1!(5 - 1)!) = 5 ways
The number of ways of the selection where there must be three boys = 5×10 = 50 ways
c. Given that Maggie and Marley must be chosen, each of whom can be chosen only one way, we have;
Selection of 2 from 8 gives;
8!/(2!(8 - 2)!) = 28
Therefore, the number of ways of the selection where there must be Maggie and Marley is 28 × 1 × 1 = 28 ways
d. The probability of no boys is given by the relation;
P(k success from n trials) = [tex]_nC_r \cdot p^k \cdot (1 - p)^{n-k}[/tex]
Where n = number of persons = 10
k = Number of person in the group = 4
p = Probability of success of girls = 1/2
Therefore, we have for no boys;
[tex]\dbinom{10}{4} \times \left (\dfrac{1}{2} \right )^4\times \left (1-\dfrac{1}{2} \right )^{10-4} = \dfrac{210}{16 \times 64} = \dfrac{105}{512} = 0.205[/tex]
4a) The number of ways of selecting 9 softball players from 15 where Rylea and Emily must be included;
Given that Rylea and Emily are already selected, we have;
The number of ways of selecting 7 softball players from 13 given as follows;
13!/(7!(13 - 7)!) = 1716 ways
Given that Rylea and Emily can both be selected in only one way, the total number of ways = 1×1×1716 = 1716 ways
