Suppose $3$ bulbs are selected at random from a lot. Each bulb is tested and classified as defective $(D)$ or non-defective $(N)$ . The sample space of this experiment is:

{D D D, D D N, D N D, N D D, D N N, N D N, N N D, N N N}

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{D D N, D N D, N D D, D N N, N D N, N N D}

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{D N N, N D N, N N D, N N N}

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{D D D, N N N}

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Solution

The correct option is A ${D D D, D D N, D N D, N D D, D N N, N D N, N N D, N N N}$
From the question,
$D$ denotes the event of bulb is defective and $N$ denotes the event of non-defective bulbs .
Then, total number of elements in sample space $= 2 \times 2 \times 2 = 8$
Thus, sample space is obtained as: $S = {D D D, D D N, D N D, N D D, D N N, N D N, N N D, N N N}$

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Sample Space

Standard XII Mathematics

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Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). The sample space of this experiment is:

The correct option is A {DDD,DDN,DND,NDD,DNN,NDN,NND,NNN}From the question, D denotes the event of bulb is defective and N denotes the event of non-defective bulbs . Then, total number of elements in sample space =2×2×2=8 Thus, sample space is obtained as: S={DDD,DDN,DND,NDD,DNN,NDN,NND,NNN}

Suppose $3$ bulbs are selected at random from a lot. Each bulb is tested and classified as defective $(D)$ or non-defective $(N)$ . The sample space of this experiment is: