A fair coin is tossed four times and the random variable X is the number of heads in the first three tosses an

Question

A fair coin is tossed four times and the random variable X is the number of heads in the first three tosses and the random variable Y is the number of heads in the last three tosses.
a) Construct a joint probability distribution of X and Y.
b) If there is one head in the last three tosses, find the conditional distribution of X.

Answer 1

Possible .X....Y
outcome
HHHH......3....3
HHHT......3....2
HHTH......2....2
HTHH......2....2
HHTT......2....1
HTHT......2....1
HTTH......1....1
HTTT......1....0
THHH......2....3
THHT......2....2
THTH......1....2
TTHH......1....2
THTT......1....1
TTHT......1....1
TTTH......0....1
TTTT......0....0

Hence,
P(X=3,Y=3)=1/16
P(X=3,Y=2)=1/16
P(X=3,Y=1)=0
P(X=3,Y=0)=0
P(X=2,Y=3)=1/16
P(X=2,Y=2)=3/16
P(X=2,Y=1)=2/16
P(X=2,Y=0)=0
P(X=1,Y=3)=0
P(X=1,Y=2)=2/16
P(X=1,Y=1)=3/16
P(X=1,Y=0)=1/16
P(X=0,Y=3)=0
P(X=0,Y=2)=0
P(X=0,Y=1)=1/16
P(X=0,Y=0)=1/16

.................JOINT PROBABILITY DISTRIBUTION OF x & y

........0.1....2.3.....|P(X=x)
...\Y|__________|
..X\.|..........................|
..__|..........................|
....0|1/16.1/16.0.0.....|2/16
......|..........................|
....1|1/16.3/16.2/16.0|2/16
......|..........................|
....2|0.2/16.3/16.1/16|6/16
......|..........................|
....3|0....0..1/16..1/16|2/16
......|..........................|
P(Y=y)|2/16.6/16.6/16.2/16|1

fOR 2nd part we have to find
P(X=3|Y=1)=P(X=3,Y=1)/P(Y=1)=0
P(X=2|Y=1)=P(X=2,Y=1)/P(Y=1)=(2/16)/(6/16)=1/3
P(X=1|Y=1)=P(X=1,Y=1)/P(Y=1)=(3/16)/(6/16)=1/2
P(X=0|Y=1)=P(X=0,Y=1)/P(Y=1)=(1/16)/(6/16)=1/6
Hence
conditonal probability distritribution of X given Y=1
......X...|0.1.2.3..|TOTAL
.....___|_______|_____
P(X=x)|0.1/3.1/2.1/6.|1

Forget about those (.........)s. They are provided so that the tables do not collapse.

Answer 2

First you need to list all the possible outcomes, let 1 be head and 0 be considered tails, then the next entry is X followed by Y

1111 X=3 Y=3
1110 X=3 Y =2
1101 X=2 Y=2
1100 X=2 Y=1
1011 X=1 Y=2
1010 X=2 Y=1
1001 X=1 Y=1
1000 X=1 Y=0
0111 X=2 Y=3
0110 X=2 Y=2
0101 X=1 Y=2
0100 X=1 Y=1
0011 X=1 Y=2
0010 X=1 Y=1
0001 X=0 Y=1
0000 X=O Y=O

Joint distribution in 1/16ths:
x:0 1 2 3 along the horizontal, y along the verticle
y|
0|1100
1|1320
2|0321
3|0011

b) If there is one head in the last three tosses find the conditional distribution of X.
We are dealing with a reduced sample space when we have conditional distributions. So reading from the row where Y = 1 we have (note that P(Y=1) = 6/16:
P(X=0|Y=1) = (1/16)/(6/16) = 1/6
P(X=1|Y=1) = (3/16)/(6/16) = 1/2
P(X=2|Y=1) = 2/16 / 6/16 = 1/3
P(X=3|Y=1) = o
Note total is equal to 1

Please check for errors.

Answer 3

1111 X=3 Y=3
1110 X=3 Y =2
1101 X=2 Y=2
1100 X=2 Y=1
1011 X=1 Y=2
1010 X=2 Y=1
1001 X=1 Y=1
1000 X=1 Y=0

Answer 4

to start flipping a coin has 2 consequences: heads or tails. a trustworthy coin skill that when you turn it the two area has an equivalent possibility of being an consequence so: 50% heads 50% tails. That being suggested you may theoretically assume to instruct a heads 0.5 of the 5 circumstances so 5/2= 2.5 so between 2 and 3

Answer 5

1st & last 3 tosses both will take on values
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
The distribution for both X and Y is
0, 1*0 = 0
1, 3*1 = 3
2, 3*2 = 6
3, 1*3 = 3
------------
... 8.... 12

b) 1 head in the last 3 tosses constrains the 1st 3 tosses to
HHT, HTH, THT, TTH
The conditional distribution, then, is
0, 0
1, 2*1 = 2
2, 2*2 = 4
------------
... 4...... 6

Answer 6

dude, that's really hard. check out the wikipedia article below for some direction. Good luck!