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
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.
1 answer