ProbabilityHard
Question
From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is :-
Options
A.$\frac{7}{10^{7}}$
B.$\frac{81}{10^{8}}$
C.$\frac{67}{10^{8}}$
D.$\frac{73}{10^{8}}$
Solution
10 defective & 90 non-defective
Req. probability $= (7$ def 1 fair $)$ or ( 8 defective)
Req. probability $= \frac{\left( 10^{7} \times 90 \right) \times 8 + 10^{8}}{100^{8}}$
$$= \frac{72 \times 10^{8} + 10^{8}}{100^{8}} = \frac{73}{10^{8}}$$
Create a free account to view solution
View Solution FreeMore Probability Questions
Three numbers are chosen at random without replacement from {1, 2, 3, ..... , 8}. The probability that their minimum is ...Two dice are tossed 6 times. Then the probability that sum 7 will show an exactly four of the tosses is-...A biased die is tossed and the respective probabilities for various faces to turn up are- If an even face has turned up,...The probability distribution of a random variable X is given below: $$X$$ 4 k $$\frac{30}{7}k$$ $$\frac{32}{7}k$$ $$\fra...A bag contains (n + 1) coins. It is known that one of these coins has a head on both sides, whereas the other coins are ...