Given: A box contains 100bulbs, 20 of which are defective.
By using the formula,
P (E) = \( \dfrac{favourable\; outcomes }{total\; possible\; outcomes}\)
Ten bulbs are drawn at random for inspection,
Total possible outcomes are 100C10
n (S) = 100C10
(i) Let E be the event that all ten bulbs are defective
n (E) = 20C10
P (E) = \( \dfrac{ n (E) }{ n (S)}\)
= \( \dfrac{ ^{20}{C}_{10}}{^{100}C_{10}}\)
(ii) Let E be the event that all ten good bulbs are selected
n (E) = 80C10
P (E) =\( \dfrac{ n (E) }{ n (S)}\)
= \(\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)
(iii) Let E be the event that at least one bulb is defective
E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs
Let E′ be the event that none of the bulb is defective
n (E′) = 80C10
P (E′) = \( \dfrac{ n (E') }{ n (S)}\)
= \( \dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)
So, P (E) = 1 – P (E′)
=\( 1-\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)
(iv) Let E be the event that none of the selected bulb is defective
n (E) = 80C10
P (E) = \( \dfrac{ n (E) }{ n (S)}\)
= \( \dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}\)
Answered by Aaryan | 1 year agoOne number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?
A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.
A die is thrown twice. What is the probability that at least one of the two throws come up with the number 3?
A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?