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Probability

Long Answer Type

921. Given three identical boxes I, II and III, each containing two coins. In box I. both coins are gold coins, in box. II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin, if the coin is of gold, what is the probability that the other coin in the box is also of gold?

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922. In a tape-recorder factory, three machines A, B, C produce 50%, 30%, 20% of the total production. The percentage of the defective output of these machines arc 3%, 4% and 5%. A tape recorder is selected at random and is found to be defective. Find the probability that the tape recorder was produced by machine A.

96 Views

Short Answer Type

923. In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

294 Views

924. In a bulb factory, machines A, B and C manufacture 60%, 30% and 10% bulbs respectively. 1%, 2% and 3% of the bulbs produced respectively by A, B and C are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by the machine A.

190 Views

Long Answer Type

925. An insurance company insured 2000 scooter drivers. 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

106 Views

Short Answer Type

926. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

86 Views

Long Answer Type

927. A doctor is to visit a patient. From the past experience it is known that the probabilities of the doctor coming by train, bus, scooter or taxi are

1 over 10 comma space 1 fifth comma space 3 over 10 space and space 2 over 5

respectively. The probabilities that he will be late are

1 fourth comma space 1 third space and space 1 over 12

if he comes by train, bus or scooter respectively but by taxi he will not be late. When he arrives he is late What is the probability that he came by bus?

81 Views

928. A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will Come by train, bus, scooter or by other means of transport are respectively

3 over 10 comma space 1 fifth comma space 1 over 10 space and space 2 over 5.

The probability that he will be late are

1 fourth comma space 1 third space and space 1 over 12 comma space

if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train ?

71 Views

Short Answer Type

929. In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is

1 third

and the probability that he copies the answer is

1 over 6.

The probability that his answer is correct (given that he copied it) is

1 over 8.

Find the probability that he knew the answer to the question (given that he correctly answered it).

584 Views

Long Answer Type

930. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA are visible. Find the probability that the letter has come from CALCUTTA.

Let E₁ be the event that the letter has come from TATANAGAR and E₂ be the event that it has come from CALCUTTA.
Let A denote the event that the two consecutive letters on the envelope are TA.
∴ E₁ and E₂ are mutually exclusive and exhaustive events
$straight P left parenthesis straight E subscript 1 right parenthesis space equals space 1 half comma space space straight P left parenthesis straight E subscript 2 right parenthesis space equals space 1 half$
P(A/E₁) = Probability that the consecutive letters TA visible on the envelope belong to TATANAGAR.
$equals space 2 over 8 space equals 1 fourth space space space space space space space space space space space space space space space space space space space open curly brackets table row cell There space are space 8 space consecutive space letters space namely end cell row cell TA comma space AT comma space TA comma space AN comma space NA comma space AG comma space GA comma space AR end cell row cell out space of space which space 2 space cases space are space favourable end cell end table space close curly brackets$

Similarly P(A/E₂) = Probability that the consecutive letters TA are visible belong to CALCUTTA
$equals space 1 over 7 space space space space space space space space space space space space space space space space space space space space space space space open curly brackets table row cell There space are space seven space consecutive space letters end cell row cell CA comma space AL comma space LC comma space CU comma space UT comma space TT comma space TA end cell row cell out space of space which space 1 space case space is space favourable end cell end table close curly brackets$

$therefore space space space straight P left parenthesis straight E subscript 2 divided by straight A right parenthesis space equals space fraction numerator straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis. space straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis. space straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis end fraction$

$equals space fraction numerator begin display style 1 half end style cross times begin display style 1 over 7 end style over denominator begin display style 1 half end style cross times begin display style 1 fourth end style plus begin display style 1 half end style cross times begin display style 1 over 7 end style end fraction space equals fraction numerator begin display style 1 over 7 end style over denominator begin display style 1 fourth end style plus begin display style 1 over 7 end style end fraction space equals fraction numerator begin display style 1 over 7 end style over denominator begin display style fraction numerator 7 plus 4 over denominator 28 end fraction end style end fraction space equals 1 over 7 cross times 28 over 11 space equals 4 over 11$

3097 Views