Chapter 13 Probability Ex 13.3 Solutions
Question - 11 : - A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas 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?
Answer - 11 : -
Let E1 bethe event of time consumed by machine A, E2 be the event oftime consumed by machine B and E3 be the event of time consumedby machine C. Let X be the event of producing defective items.
Then P (E1)= 50% = 50/100 = ½
P (E2) =30% = 30/100 = 3/10
P (E3) =20% = 20/100 = 1/5
As we a headed coinhas head on both sides so it will shows head.
Also P (X|E1)= P (defective item produced by A) = 1% = 1/100
And P (X|E2)= P (defective item produced by B) = 5% = 5/100
And P (X|E3)= P (defective item produced by C) = 7% = 7/100
Now the probabilitythat item produced by machine A, being given that defective item is produced,is P (E1|A).
By using Bayes’theorem, we have
P (E1|A) =5/34
Question - 12 : - A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer - 12 : -
Let E1 be theevent that the drawn card is a diamond, E2 be the event that the drawncard is not a diamond and A be the event that the card is lost.
As we know, out of 52cards, 13 cards are diamond and 39 cards are not diamond.
Then P (E1)= 13/52 and P (E2) = 39/52
Now, when a diamondcard is lost then there are 12 diamond cards out of total 51 cards.
Two diamond cards canbe drawn out of 12 diamond cards in 12C2 ways.
Similarly, two diamondcards can be drawn out of total 51 cards in 51C2 ways.
Then probability ofgetting two cards, when one diamond card is lost, is P (A|E1).
Also P (A|E1)=12C2 / 51C2
Now, when not adiamond card is lost then there are 13 diamond cards out of total 51 cards.
Two diamond cards canbe drawn out of 13 diamond cards in 13C2 ways.
Similarly, two diamondcards can be drawn out of total 51 cards in 51C2 ways.
Then probability ofgetting two cards, when card is lost which is not diamond, is P (A|E2).
Also P (A|E2)=13C2 / 51C2
Question - 13 : - Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
A. 4/5
B. 1/2
C. 1/5
D. 2/5
Answer - 13 : -
A. 4/5
Explanation:
Let E1 bethe event that A speaks truth, E2 be the event that A lies andX be the event that it appears head.
Therefore, P (E1)= 4/5
As E1 andE2 are the events which are complimentary to each other.
Then P (E1)+ P (E2) = 1
⇒ P (E2)= 1 – P (E1)
⇒ P (E2) =1 – 4/5 = 1/5
If a coin is tossed itmay show head or tail.
Hence the probabilityof getting head is 1/2 whether A speaks a truth or A lies.
P (X|E1) =P (X|E2) = ½
Now the probabilitythat actually there was head, give that A speaks a truth is P (E1|X).
By using Bayes’theorem, we have
Question - 14 : - If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
A. P (A| B) = P (B)/ P (A)
B. P(A|B) < P(A)
C. P(A|B) ≥ P(A)
D. None of these
Answer - 14 : -
C. P (A|B) ≥ P (A)
Explanation:
