Find the probability distribution of the number of heads in the simultaneous tosses of three coins (or number of heads in three tosses of a coin). Also find the mean number of heads.
Let X be the number of heads in simultaneous tosses of three coins.
X can take the values X = 0,1,2,3
Sample space = { HHH,HHT,HTH,THH,HTT,THT,TTH,TTT }
n(S) = 8
P[X=0] =P[no head] = P[{ TTT }] = ( 1/8 )
P[X=1] =P[one head] = P[{ HTT,THT,TTH }] = ( 3/8 )
P[X=2] =P[two heads ] = P[{ HHT,THH,HTH }] = ( 3/8 )
P[X=3] =P[three heads] = P[{ HHH}] = ( 1/8 )
Mean =E[X] = 0( 1/8 ) + 1(3/8 ) + 2 ( 3/8 ) + 3( 1/8 ) = (12/8) Mean = (3/2) = 1.5
=========================================================================
A
random variable X has the following probability distribution:
Find
(i) k (ii) P(X <
3)(iii) P(X > 6) (iv) P(0 < X < 3)
Sum
of all values of P[X] = 1
0
+ k + 2k
+ 2k + 3k
+ k2 + 2k2
+ 7k2+k
= 1
10k2
+ 9k = 1
10k2
+ 9k – 1=0
10k2
+ 10k - 1k – 1=0
10k(k+1)
-1 (k+1) = 0
(10k-1)
(k+1) = 0
k
= (1/10) or k = (-1)
Since
probability cannot be negative , we reject k = (-1)
therefore
k = (1/10)
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Two
numbers are selected at random (without replacement) from the first
six positive integers. Let X denote the larger of the two numbers
obtained. Find E(X).
Let
X be the larger of the two numbers.
X
can take the values 2,3,4,5,6
If
S is the sample space
S
= { (1,2), (1,3),(1,4),(1,5),(1,6),
(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),
(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,1),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,1),(6,2),(6,3),(6,4),(6,5)}
n(S)
= 6*5 =30 [since replacement is not allowed]
P[X=2]
= n[{(1,2),(2,1)}] / 30 = (2/30)
P[X=3]
= n[{(1,3),(2,3),(3,1),(3,2)}] / 30 = (4/30)
P[X=4]
= n[{(1,4),(2,4),(3,4),(4,1),(4,1),(4,2),(4,3)}] / 30 = (6/30)
P[X=5]
= n[{(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4)}] / 30 = (8/30)
P[X=6]
=
n[{(6,1),(6,1),(6,2),(6,3),(6,4),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6)}]
= (10/30)
Mean
= E[X] = 2 (2/30) + 3(4/30) + 4(6/30) + 5 (8/30) + 6 (10/30) =
Mean
= (140/30) = (14/3)
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