I CANT DO THESE TWO QUESTIONS PLEASE ANYONE SAVE ME 😭 (1 Viewer)

[Tryingtodowell]

 

[liamkk112]

for 1) (assuming coin is 50/50), use P(H) = C(9,H) * (0.5)^H * (0.5)^(9-H), H is the number of heads

2a)there are C(10,6) ways to choose 6 students from the group of 10, not considering order (this is basically just the definition of the choose function)
b) now 2 of the six slots are filled, and again not considering order there is 1 way to do this. now there is 8 remaining students, so the total number of ways is 1x C(8,4) = C(8,4)
c) now you're just picking from a group of 9 students, so there are C(9,6) ways
d) i'm assuming this question means that when you pick the 6 students, you can't pick both A and B. there are two cases: one where you pick 1 of A and B, and one where you pick neither. for the first case, you have 2 options to pick the first student (A or B), and then C(8,5) ways to pick the other five studnets, giving 2xC(8,5) ways
for the second case, you have only 8 students to pick from, so there are C(8,6) ways
so in total, there are 2xC(8,5) + C(8,6) ways

hopefully thats right

 

[cossine]

View attachment 43328

pascal triangle represents the values of nCr. Once you understand how it is defined the questions are straightforward.

You need to then make use of the formula probability = num_permutations*probability_of_single_events. Make sure to understand why this formula holds.