Please help me anybody
Abby Garland is parked at a mile marker on an east -west country road.She decides to toss a fair coin 10 times ,each time driving 1 mile east if it lands heads up and 1 mile west if it lands up tails up. The term random walk applies to this process even though Abby drives rather than walks. it is a simplifiedmodel ofBrownian motion.Find the probability that Abby's walk will end as described.
A 10 miles east of start.( answer 1/1024)
B 6 miles east of start
C.6 miles west of start (answer 45/1024)
D.5 miles west of the start
E 2 miles east of start(answr210/1024)
F.at least 2 miles east of the start
G.at least 2 miles from the start (772/10240)
H. Exactly at the start
Please give me a detail explanation of how to start the problem.
Compute the following probabilities.
A. 10 heads and zero tails
P = (1/2)^10 = 1/1024
B. 8 heads and 2 tails
C. 2 heads and 8 tails
P = (1/2)^10*10!/(8!*2!] = 45/1024
D. zero (not possible; there must be ab even number of place changes)
E. 6 tails and 4 heads (1/2)^10*10!/7!*3!)= (1/1024)*8[10!/(3!*7!)]=120/1024
F. 6,7,8, 9 or 10 heads
G. anything but 5 heads and 5 tails
Your denominator should be 1024, not 10240
H. 5 heads and 5 tails
P = (1/2)^5*(1/2)^5*[10!/5!]^2
To find the probability of Abby's walk ending as described, we need to analyze the possible outcomes of tossing a fair coin 10 times and determine the number of outcomes that result in each scenario.
To get started, let's determine the number of possible outcomes when tossing a fair coin 10 times. Since each toss has 2 possible outcomes (heads or tails), the total number of outcomes will be 2^10 = 1024.
Now, let's break down each scenario and calculate the number of outcomes that lead to them:
A) 10 miles east of start:
For Abby to end up 10 miles east of the start, she must get heads for all 10 coin tosses. Since each toss has a 1/2 probability of landing heads up, the probability of getting heads 10 times in a row is (1/2)^10 = 1/1024. Therefore, the number of outcomes leading to this scenario is 1.
B) 6 miles east of start:
To be 6 miles east of the start, Abby needs to have more heads than tails by 4 tosses. We can approach this problem by considering the number of ways Abby can get 4, 5, 6, 7, 8, 9, or 10 heads.
For 4 heads, Abby must have 6 tails. The number of ways to arrange 4 heads and 6 tails is given by the binomial coefficient (10 choose 4), which is equal to 10! / (4! * (10-4)!) = 210.
Similarly, for 5, 6, 7, 8, 9, and 10 heads, the number of ways to arrange the heads and tails can be found using the binomial coefficients:
(10 choose 5) = 10! / (5! * (10-5)!) = 252,
(10 choose 6) = 210,
(10 choose 7) = 120,
(10 choose 8) = 45,
(10 choose 9) = 10, and
(10 choose 10) = 1.
The probability of each event occurring is (1/2)^10 since each toss is independent. So, we can calculate the sum of the probabilities for 4 to 10 heads as follows:
(1/2)^10 * (210 + 252 + 210 + 120 + 45 + 10 + 1) = 45/1024.
Thus, the number of outcomes leading to Abby being 6 miles east of the start is 45.
C) 6 miles west of start:
Similarly to scenario B, Abby needs to have more tails than heads by 4 tosses. This will be the mirror image of scenario B, so the number of outcomes leading to this scenario will also be 45/1024.
D) 5 miles west of start:
To be 5 miles west of the start, Abby needs to have more tails than heads by 5 tosses. As with scenario B, we consider the number of ways Abby can get 5, 6, 7, 8, 9, or 10 tails.
By applying the same logic as before, we find that the number of outcomes leading to this scenario is:
(1/2)^10 * (252 + 210 + 120 + 45 + 10 + 1) = 383/1024.
E) 2 miles east of start:
To be 2 miles east of the start, Abby needs to have more heads than tails by 8 tosses. We calculate the number of outcomes leading to this scenario in a similar manner as before:
(1/2)^10 * (45 + 10 + 1) = 56/1024 = 7/128.
Thus, the number of outcomes leading to Abby being 2 miles east of the start is 7.
F) At least 2 miles east of start:
This scenario includes Abby being either 2, 3, 4, 5, 6, 7, 8, 9, or 10 miles east of the start. We sum the number of outcomes for each individual scenario:
7 + 28 + 84 + 210 + 210 + 120 + 45 + 10 + 1 = 715.
G) At least 2 miles from the start:
This scenario considers Abby being either 2 miles east or 2 miles west of the start. The number of outcomes leading to this scenario is the sum of the outcomes for being 2 miles east of start (7) and for being 2 miles west of start (45):
7 + 45 = 52.
H) Exactly at the start:
In this scenario, Abby would need to have an equal number of heads and tails (5 of each) in 10 tosses. The number of outcomes leading to this scenario is:
(1/2)^10 * (10 choose 5) = 252/1024.
Therefore, the answer choices are:
A) 1/1024
B) 6/1024 = 3/512
C) 45/1024
D) 383/1024
E) 7/128
F) 715/1024
G) 52/1024 = 13/256
H) 252/1024 = 63/256
Hence, the correct answers are C) 45/1024 and E) 7/128.