An experiment consists of tossing a coin 14 times.

(a) How many different outcomes are possible?
(b) How many different outcomes have exactly 9 heads?
(c) How many different outcomes have at least 2 heads?
(d) How many different outcomes have at most 10 heads?

To solve these questions, we can use the concept of combinations and the formula for calculating combinations, which is:

nCr = n! / (r! * (n-r)!)

Where n represents the total number of trials, and r represents the number of successful outcomes.

(a) To find the number of different outcomes possible when tossing a coin 14 times, we can use the formula for calculating combinations:

nCr = 14C14 = 14! / (14! * (14-14)!) = 1

Therefore, there is only one possible outcome when tossing a coin 14 times.

(b) To find the number of different outcomes that have exactly 9 heads, we can use the formula for calculating combinations:

nCr = 14C9 = 14! / (9! * (14-9)!) = 2002

Therefore, there are 2002 different outcomes that have exactly 9 heads.

(c) To find the number of different outcomes that have at least 2 heads, we need to calculate the combinations for 2, 3, 4, ..., up to 14 heads, and then sum them up. Alternatively, we can find the total number of outcomes and subtract the ones with 0 or 1 head.

Total outcomes possible = 2^14 (since each toss can have 2 possible outcomes - heads or tails)

Number of outcomes with 0 heads = 1 (tails only)
Number of outcomes with 1 head = 14C1 = 14 (only 1 of the 14 tosses is a head)

Therefore, the number of outcomes with at least 2 heads = Total outcomes - Outcomes with 0 or 1 head = 2^14 - (1 + 14) = 16384 - 15 = 16369

Therefore, there are 16369 different outcomes that have at least 2 heads.

(d) To find the number of different outcomes with at most 10 heads, we need to calculate the combinations for 0, 1, 2, ..., up to 10 heads, and then sum them up.

Number of outcomes with 0 heads = 1 (tails only)
Number of outcomes with 1 head = 14C1 = 14
Number of outcomes with 2 heads = 14C2 = 91
Number of outcomes with 3 heads = 14C3 = 364
Number of outcomes with 4 heads = 14C4 = 1001
Number of outcomes with 5 heads = 14C5 = 2002
Number of outcomes with 6 heads = 14C6 = 3003
Number of outcomes with 7 heads = 14C7 = 3432
Number of outcomes with 8 heads = 14C8 = 3003
Number of outcomes with 9 heads = 14C9 = 2002
Number of outcomes with 10 heads = 14C10 = 91

Therefore, the number of outcomes with at most 10 heads = Sum of all the above = 1 + 14 + 91 + 364 + 1001 + 2002 + 3003 + 3432 + 3003 + 2002 + 91 = 11175

Therefore, there are 11175 different outcomes that have at most 10 heads.

To answer these questions, we will use the concept of permutations and combinations.

(a) To find the number of different outcomes possible when tossing a coin 14 times, we need to consider each toss as an independent event. Since there are two possible outcomes (heads or tails) for each toss, we can use the formula for the total number of outcomes in multiple events:

Total outcomes = (Number of outcomes for 1st event) x (Number of outcomes for 2nd event) x ... x (Number of outcomes for 14th event)

In this case, the number of outcomes for each event is 2 (heads or tails). Therefore, the total number of different outcomes possible when tossing a coin 14 times is 2^14 = 16,384.

(b) To find the number of different outcomes that have exactly 9 heads, we need to choose 9 out of the 14 tosses to be heads and the remaining 5 tosses to be tails. We can use the combination formula:

Number of outcomes with exactly 9 heads = C(14, 9) = 14! / (9! * (14-9)!) = 2002

(c) To find the number of different outcomes that have at least 2 heads, we need to consider two cases: one with exactly 2 heads and the other with more than 2 heads.

For the case with exactly 2 heads, we need to choose 2 out of the 14 tosses to be heads and the remaining 12 tosses to be tails:

Number of outcomes with exactly 2 heads = C(14, 2) = 14! / (2! * (14-2)!) = 91

For the case with more than 2 heads, we can use the concept of complement:

Number of outcomes with at least 2 heads = Total outcomes - Number of outcomes with no heads or 1 head

Number of outcomes with no heads = C(14, 0) = 1
Number of outcomes with 1 head = C(14, 1) = 14

Number of outcomes with at least 2 heads = 16,384 - 1 - 14 = 16,369

(d) To find the number of different outcomes that have at most 10 heads, we need to consider two cases: one with 0 to 10 heads.

Number of outcomes with 0 heads = C(14, 0) = 1
Number of outcomes with 1 head = C(14, 1) = 14
Number of outcomes with 2 heads = C(14, 2) = 91
Number of outcomes with 3 heads = C(14, 3) = 364
Number of outcomes with 4 heads = C(14, 4) = 1001
Number of outcomes with 5 heads = C(14, 5) = 2002
Number of outcomes with 6 heads = C(14, 6) = 3003
Number of outcomes with 7 heads = C(14, 7) = 3432
Number of outcomes with 8 heads = C(14, 8) = 3003
Number of outcomes with 9 heads = C(14, 9) = 2002
Number of outcomes with 10 heads = C(14, 10) = 1001

Number of outcomes with at most 10 heads = Sum of outcomes with 0 to 10 heads = 1 + 14 + 91 + 364 + 1001 + 2002 + 3003 + 3432 + 3003 + 2002 + 1001 = 13,568

a) each turn could be 2 ways, so 2^14 = 16384

b) this is the same as asking, "in how many ways can we arrange 9 H's and 5 T's, the H's and T's are indistinguishable.

number of ways = 14!/(9!5!) = 2002

c)at least two heads implies we don't want 0 heads, or 1 head, let's find those two
0 heads ---> 1 way
1 head -----> 14!/13! = 14
so at least 2 heads = 16384 - 14 - 1 = 16369

d) so we don't want 11 heads, 12 heads, 13 heads and 14 heads
which are 14!/(11!3!) + 14!/(12!2!) + 14!/(13!1!) + 1
= 364 + 91 + 14 + 1 = 470

so at most 10 heads = 16384 - 470 = 15914