Drag the tiles to the correct boxes to complete the pairs. A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability.​

The selection of r objects out of n is done in

many ways.

The total number of selections 10 that we can make from 6+7=13 students is  

thus, the sample space of the experiment is 286

A.  

"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

P(all 6 girls chosen)=35/286=0.12

B.

"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is  

so the probability is 20/286=0.07

C.

"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.  

the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.

"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in  

many ways,

so the probability is 126/286=0.44

Mark brainliest

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.

50

step-by-step explanation: