A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, then what is the probability that there will be atleast one defective watch ?
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A bag contains 6 red, 3 black and 6 white balls. A ball is selected at random from the bag, what is the probability that the selected ball is not red?
Correct option is: A) \(\frac{3}{5}\) There are 6 red, 3 black and 6 white balls. \(\therefore\) Total number of balls = 6 + 3 + 6 = 15. Number of balls which are not red =...
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A factory produces bulbs. The probability that any one bulb is defective is `1/50` and they are packed in 10 boxes. From a single box, find the probab
Let X is the random variable which denotes that a bulb is defective. Also, n=10,`p=1/50` and q=`49/50` and P(X=r)=`""^(n)C_(r)p^(r )q^(n-r)` (i) None of the bulbs is defective i.e.,r=0 `thereforep(X=r)=P((0))=""^(10)C_(0)(1/50)^(0)(49/50)^(10-0)=(49/50)^(10)` (ii)...
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There are 5 cards numbered 1 to 5 , one number on one number on one card . Two cards are drawn at random without replacement . Let X denotes the sum o
Here, S={(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)} `rArrn(S)=20` Let random variable variable be X which denotes the sum of the numbers on two cards drawn. X=3,4,5,6,7,8,9 At X=3,P(X)=`2/20=1/10` At X=4,P(X)`=2/10=1/10` At X =5,P(X)=`4/20=1/5` At X=6,...
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A flashlight has 8 batteries out of which 3 are dead. If two batteries selected without replacement and tested, then probability that both are dead is
Required probability=`P_(D)cdotP_(D)=3/8cdot2/7=3/28`
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If a box has 100 pens of which 10 are defective, then what is the probability that out of a sample of 5 pens drawn one by one with replacement atmost
Here, n=5,p`=10/100=1/10andq=9/10` `rle1` `rArrr=0,1` Also, `P(X=r)=""^(n)C_(r)p^(r )q^(n-r)` `therefore P(X=r)=P(r=0)+P(r=1)` `=""^(5)C_(0)(1/10)^(0)(9/10)^(4)` `=(9/10)^(5)+5cdot1/10cdot(9/10)^(4)` `(9/10)^(5)+1/2(9/10)^(4)`
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A three-digit number is selected at random from the set of all three-digit numbers. The probability that the number selected has all the three digits
Correct Answer - D Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...
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Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`
To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...
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Suppose A and B shoot independently until each hits his target. They have probabilities `3/5` and `5/7` of hitting the target at each shot. The probab
Let Xbe the number of times A shoots at the target to hit it for the first time and Y be the number of times B shoots at the target...
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Ina certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes. If a person selected at random
Correct Answer - B We have, `P(A)=40/100,P(B)=25/100and P(AnnB)=15/100` So, `P((B)/(A))=(P(AnnB))/(P(A))=(15//100)/(40//100)=3/8`
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A lot contains 50 defective and 50 non-defectivebulbs. Two bulbs are drawn at random one at a time withreplacementevents A, and as first bul. The B C
Correct Answer - A::B::C Let X = defective and Y = non-defective. Then all possible outcomes are `{XX,XY,YX,YY}` Aslo, `P(XX)=(50)/(100)xx(50)/(100)=1/4` `P(XY)=(50)/(100)xx(50)/(100)=1/4` `P(YX)=(50)/(100)xx(50)/(100)=1/4` `P(YY)=(50)/(100)xx(50)/(100)=1/4` Here, `A=(XX)uu(XY),B=(XY)uu(YY),C=(XX)uu(YY)` `thereforeP(A)=P(XX)+P(XY)=1/4+1/4=1/2` `thereforeP(B)=P(XY)+P(YX)=1/4+1/4=1/2` Now, `P(AxxB)=P(XY)=1/4=P(A)xxP(B)` Thus, A...
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