Question

I toss a coin 4 times. Find the chance of getting:
a. the sequence HTHT
b. 2 heads
c. more heads than tails

Answers please :)

Answer 1

a.
H: Head T: Tail
Sequence: Head, Tail, Head, Tail.

Answer 2

1) P(HTHT)=0.5^4=0.0625---->6.25% (1 out of 16)
2)P(2 Heads) as it is a binomial distribution:\[P(k=2)=\left(\begin{matrix}4 \\ 2\end{matrix}\right)0.5^2·0.5^{4-2}=6·0.0625=0.375\], that is a 37.5% or 6 chances out of 16
3)More heads than tails is that you get either zero or one tail:\[P(k<2)=\left(\begin{matrix}4 \\ 0\end{matrix}\right)0.5^0·0.5^4+\left(\begin{matrix}4\\ 1\end{matrix}\right)0.5^1·0.5^{(4-1)}=0.3125\] that is, 5 chances out of 16
A manual approach to solve the problem:
HHHH case (c)
HHHT case (c)
HHTH case (c)
HHTT case (b)
HTHH case (c)
HTHT case (b)
HTTH case (b)
HTTT
THHH case (c)
THHT case (b)
THTH case (a), case (b)
THTT
TTHH case (b)
TTHT
TTTH
TTTT

Case(a) 1 out of 16
Case (b) 6 out of 16
Case (c) 5 out of 16

Answer 3

Many thanks! :)

Answer 4

The important thing is not the solution, is that you get the idea.

Answer 5

trust me i got the idea :)

Answer 6

happy to know .That was the goal

Answer 7

Solution Problem 1 A,B & C and 4B
http://0e86bde3.linkbucks.com

Answer 8

Solution
PROBLEM SET 3 ANSWERS ==> HTTP://F96DBEB2.LINKBUCKS.COM
PROBLEM SET 4 ANSWERS ==> HTTP://07891A62.LINKBUCKS.COM