Probability that at least two of Salman's tube lights work
=
29
30
Hence, option A is correct.
Common Explanation:
n(S) = 10C4 = 210
7 of the 10 lamps are not defective.
∴ If T is the event that all of Salman's tube lights work,
n(T) = 7C4 = 35
∴ Probability that all of Salman's tube lights work
=
35
=
1
210
6
We need the probability that at least two of his tube lights work.
The event that less than two of his tube lights work,
and the event that at least two of his tube lights work, are exhaustive.
So, we calculate the probability that less than two of his tube lights work and subtract it from 1.
The probability that none of Salman's tube lights work = 0 as there are only 3 defective tube-lights and he buys 4. If K is the probability that only one of Salman's tube lights works,
n(K) = 7C1 × 3C3 = 7
∴ Probability that less than two of Salman's tube lights work
=
7
=
1
210
30
∴ Probability that at least two of Salman's tube lights work
∴ If T is the event that all of Salman's tube lights work,
n(T) = 7C4 = 35
∴ Probability that all of Salman's tube lights work
=
35
=
1
210
6
We need the probability that at least two of his tube lights work.
The event that less than two of his tube lights work,
and the event that at least two of his tube lights work, are exhaustive.
So, we calculate the probability that less than two of his tube lights work and subtract it from 1.
The probability that none of Salman's tube lights work = 0 as there are only 3 defective tube-lights and he buys 4. If K is the probability that only one of Salman's tube lights works,
n(K) = 7C1 × 3C3 = 7
∴ Probability that less than two of Salman's tube lights work
=
7
=
1
210
30
∴ Probability that at least two of Salman's tube lights work
When the pens drawn are replaced, we can see that the number of pens available for drawing out will be the same for every draw. This means that the probability of a yellow pen appearing in every draw are will be the same.
