# A pair of dice is

Given : A die is tossed twice and ‘Getting a number greater than 4 ’ is considered a success.

To find : probability distribution of the number of successes and mean (𝓊) and variance (σ2)

Formula used : Mean = E(X) = Variance = E(X2) - When a die is tossed 4 times,

Total possible outcomes = 62 = 36

Getting a doublet is considered as a success

The possible doublets are (1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)

Let p be the probability of success,

p = = q = 1 – p = 1 - = q = since the die is thrown 4 times, n = 4

x can take the values of 1,2,3,4

P(x) = nCx P(0) = 4C0 = P(1) = 4C1 = = P(2) = 4C2 = = P(3) = 4C3 = = P(4) = 4C4 = The probability distribution table is as follows, Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5)

Mean = E(X) = 0( ) + 1( ) + 2( ) + 3( ) + 4( )

Mean = E(X) = 0 + + + + = = = Mean = E(X) =  = = E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5)

E(X2) = ( ) + ( ) + ( ) + ( ) + ( )

E(X2) = 0 + + + + = = E(X2) = 1

Variance = E(X2) - = 1 – = Variance = E(X2) - = The probability distribution table is as follows, Mean = Variance = Rate this question :

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