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 amongthe last three flips

The answers to the question are

(a) 1/4

(b) 7/16

(c) 1/16

Step-by-step explanation:

To solve the question we note that

Total number  of outcomes = 32

Probability of the event of first three flips are the same =P(F)

Probability of the event of last three flips are the same =P(L)

Total number of outcomes = 2⁵ = 32

The number of ways in which the frist three flips are the same is

F  , H, TTTHH, HHHTT, T, , TTTHT, HHHTH = 8

L:  , HHTTT, H, THTTT, , TTHHH, HTHHH, T = 8

The probability that the first and the last three flips are the same that is

F ∩ L; , = 2

Therefore P(F ∩ L) = 2/32

(a) P(F) = 8/32 =1/4 also

     P(L) = 8/32 =1/4

(b) P(LUF) =  P(L) + P(F) - P(F ∩ L) = 1/4+1/4-1/16 =7/16

(c) Let the event of at least two heads among the first three flips be H

and the event of at least two tails among the last three flips be T

Then we have

H; , T, HHHTT, HHHTH, HHTTT, HHTHH, THHTT, THHHT

= 8

T; . H, HHTTT, THTTT, THHTT, HHHTT, THHTT, HTTTH =8

Also H∩T  = TTTHH, HHTTT = 2

Therefore P(H ∩ T) = 2/32 = 1/16

.mp4 files, because, video

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%