MATH SOLVE

3 months ago

Q:
# Five independent flips of a fair coin are made. find the probability that (a) the first three flips are the same; (b) either the first three flips are the same, or the last three flips are the same; (c) there are at least two heads among the first three flips, and at least two tails among the last three flips.

Accepted Solution

A:

Answer:a) 1/4b) 1/2c) 9/64Step-by-step explanation:The possible outcome of five independent flip = 2^5 = 32a) In order to determine the probability that the first three flips are the same, we have to list out the possible outcomes. The first three flips being the same could be HHH, TTT= 2 waysThe last two flips could be HH, HT, TH, TT= 4 waysTherefore the possible ways of obtaining the first three flips as being the same in five independent flips = 2*4= 8 waysHHHHH, HHHHT, HHHTH, HHHTT, TTTHH, TTTHT, TTTTH, TTTTTTherefore the possible ways that the first three flips are the same = 8/32= 1/4b) To also determine that the last three flips are the same, we will list out the possible outcomes of the last three flips TTT, HHH= 2 waysThe first two flips could be HH, HT, TH, TT= 4waysTherefore, the possible ways to obtain the last three flips as the same in five independent flips = 2*4= 8 ways HHTTT, HTTTT, THTTT, TTTTTHHHHH, HTHHH, THHHH, TTHHHthe probability that the last three flips are the same = 8/32= 1/4Since the probability that the first three flips are the same = 1/4The probability that either the first three flips are the same or the last three flips are the same = 1/4 + 1/4= 2/4= 1/2c) To determine the probability that there are at least two heads among the first three flips, list out the possible outcomes. The first three flips with at least two heads are; HHH, THH, HHT= 3 waysThe last two flips could be HH, HT, TH, TT= 4 waysTherefore the possible ways of obtaining the first three flips with at least two heads = 3*4= 12 waysHHHHH, HHHHT, HHHTH, HHHTT, THHHH, THHHT, THHTH, THHTT, HHTHH, HHTHT, HHTTH, HHTTTThe probability that we have at least two heads among first three flips= 12/32= 3/8To also determine the probability that there are at least two tails among the last three flips, list out the possible outcomes. The last three flips with at least two tails are; TTH, THT, HTT= 3 waysThe first two flips could be HH, HT, TH, TT= 4 waysTherefore the possible ways of obtaining the last three flips with at least two tails= 3*4= 12 waysHHTTH, HTTTH, THTTH, TTTTH, HHTHT, HTTHT, THTHT, TTTHT, HHHTT, HTHTT, THHTT, TTHTTThe probability that we have at least two tails among the last three flips= 12/32= 3/8Therefore the probability that there are at least two heads among the first three flips and at least two tails among the last three flip = 3/8 * 3/8= 9/64