Q. 14

# In the items prod

(i) The probability that the item is defective = = p

The probability that the bulb will not fuse = 1 = = q

Using Bernoulli’s we have,

P(Success=x) = nCx.px.q(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability that exactly 2 defective items are,

6C2.( )2( )4

(ii) The probability that the item is defective = = p

The probability that the bulb will not fuse = 1 = = q

Using Bernoulli’s we have,

P(Success=x) = nCx.px.q(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability that not more than 2 defective items are,

6C0.( )0( )6 + 6C1.( )1( )5 + 6C2.( )2( )4 (iii) The probability that the item is defective = = p

The probability that the bulb will not fuse = 1 = = q

Using Bernoulli’s we have,

P(Success=x) = nCx.px.q(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability of at least 3 defective items are,

P(3) + P(4) + P(5) + P(6)

6C3.( )3( )3 + 6C4.( )4( )2 + 6C5.( )5( )1 + 6C6.( )6( )0 Rate this question :

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