probability distribution and the mean number of the number of heads in the simultaneous tosses of three coins

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

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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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