A box contains 20 bulbs of which 3 are defective. If 3 bulbs are chosen at random then, the probability of having at least one good bulb is: A.$\dfrac{169}{190}$ B.$\dfrac{189}{190}$ C.$\dfrac{75}{95}$ D.$\dfrac{1}{190}$
ANSWER
Verified
Hint: Here, probability of at least one good bulb is calculated as (1 – probability of zero good bulbs). Then we have to calculate the probability of zero good bulbs by calculating $\dfrac{3!}{{}^{20}{{C}_{3}}}$ where $3!$is the total number of ways to choose 3 defective bulbs and ${}^{20}{{C}_{3}}$ is the total ways to choose 3 defective bulbs from 20 bulbs.
Complete step-by-step answer: Here, we are given that a box contains 20 bulbs of which 3 are defective and 3 bulbs are chosen at random. Now, we have to find the probability of having at least one good bulb. From the given data we can write: Total number of bulbs = 20 Total number of defective bulbs = 3 Total number of good bulbs = 17 So, the probability of having at least one bulb is the complement of probability of having zero good bulbs. We know that the total probability is 1. Hence, we can write: Probability of having at least one good bulb = 1 – Probability of having zero good bulbs. Now, let us calculate the probability of having zero good bulbs. If zero good bulbs are chosen means all the 3 bulbs chosen are defective out of 20 bulbs. The total number of possible ways to choose 3 bulbs = ${}^{20}{{C}_{3}}$ The number of possible ways to choose 3 bulbs out of 3 defective bulbs = $3!$ Therefore, we can write: Probability of having zero good bulbs = $\dfrac{3!}{{}^{20}{{C}_{3}}}$ ….. (1) We know by the combination formula that: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ Now, consider ${}^{20}{{C}_{3}}$, $\begin{align} & {}^{20}{{C}_{3}}=\dfrac{20!}{3!\times \left( 20-3 \right)!} \\ & {}^{20}{{C}_{3}}=\dfrac{20!}{3!\times 17!} \\ \end{align}$ We know that $20!=17!\times 18\times 19\times 20$. Hence, we will get: $\begin{align} & {}^{20}{{C}_{3}}=\dfrac{17!\times 18\times 19\times 20}{3!\times 17!} \\ & {}^{20}{{C}_{3}}=\dfrac{17!\times 18\times 19\times 20}{1\times 2\times 3\times 17!} \\ \end{align}$ Next, by cancellation we obtain: ${}^{20}{{C}_{3}}=3\times 19\times 20$ We also know that: $\begin{align} & 3!=1\times 2\times 3 \\ & 3!=6 \\ \end{align}$ Hence, by substituting these values in equation (1) we get: Probability of having zero good bulbs = $\dfrac{6}{3\times 19\times 20}$ Next, by cancellation we obtain: Probability of having zero good bulbs = $\dfrac{1}{190}$ Hence we can write: Probability of having at least one good bulb = $1-\dfrac{1}{190}$ In the next step, by taking LCM we get: Probability of having at least one good bulb = $\dfrac{190-1}{190}$ Probability of having at least one good bulb = $\dfrac{189}{190}$ Hence, we can say that the probability of having one good bulb is $\dfrac{189}{190}$. Therefore, the correct answer for this question is option (b).
Note: Here, they have used the word at least so you have to take the compliment of the probability which will be easier to calculate. Also don’t forget that the total probability is 1.