The probability of happening an event A in one trial is 0.4. Find the probability that the event A happens at least one in three independent trials.
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Which one is not a requirement of a binomial distribution? There are 2 outcomes for each trial There is a fixed number of trials The outcomes must be
we know that, in a Binomial distribution, (i) There are 2 outcomes for each trial. (ii) There is a fixed number of trials. (iii) The probability of succes must be...
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The probability that at least one of the events `Aa n dB` occurs is 0.6. If `Aa n dB` occur simultaneously with probability 0.2, then find `P( A )+P(
It is given that `P(A uu B) = 0.6` and `P(A nn B) = 0.2`. `P(A uu B) = P(A) + P(B) - P(A nn B)` or `0.6 = P(A)...
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The chance of an event happening is the square of the chance of a second event but the odds against the first are the cube of the odds against the sec
Correct Answer - A::C Let `p_(1), p_(2)` be the chances of happening of the first and second events, respectively. Then according to the given conditions, we have `p_(1) = p_(2)^(2)` and...
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If A and B are two independent events, the probability that both A and B occur is 1/8 are the probability that neither of them occours is 3/8. Find th
We have, `P(AnnB)=1/8and P(barAnnbarB)=3/8` `thereforeP(A)P(B)=1/8and P(barA)P(barB)=3/8` `" "[therefore"A and B are independent"]` Now, `P(barAnnbarB)=3/8` `implies1-P(AnnB)=3/8` `or 1-(P(A)+P(B)-P(AnnB))=3/8` `or1-(P(A)+P(B))+1/8=3/8` `orP(A)+P(B)=3/4` The equation whose roots are P(A) and P(B) is `x^(2)-x{P(A)+P(B)}+P(A)P(B)=0` `or...
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An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.
Let p be its probability of success and q be the failure. Then p =2q. Also, p+q=1. It gives `p=2//3and q=1//3.` P {4 successes in 6 trials} `=""^(6)C_(4)p^(4)q^(2)` `=""^(6)C_(4)((2)/(3))^(4)((1)/(3))^(2)" "(1)`...
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A coin is tossed three times in succession. If `E` is the event that there are at least two heads and `F` is the event in which first throw is a head,
Correct Answer - `3//4` `P(E)=4/8=1/2{because"favorable cases are"THH, HTH, HHT, HHH]` `P(F)=4/8=1/2[because"favorable cases are"HTH, HHT,HT T, HHH]` `P(EnnF)=3/8` `P(E//F)=(P(EnnF))/(P( F))=(3//8)/(1//2)=3/4`
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A coin is tossed three times. Event A: two heads appear Event B: last should be head Then identify whether events A and B are independent or not.
Correct Answer - Not independent `P(A)=3/8and P(B)=1/2` `P(A)P(B) =3/8xx1/2=3/16` `P(Annb)=2/8=1/4neP(A)P(B)` Hence, A and B aer not independent.
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Let A,B, C be three mutually independent events. Consider the two statements `S_(1)and S_(2).` `{:(S_(1):A and B nnC "are independent.",),(S_(2):A and
Correct Answer - A We are given that `P(AnnB)=P(A)P(B)` `P(BnnC)=P(B)P(C)` `P(CnnA)=P(C)P(A)` `P(AnnBnnC)=P(A)(B)P(C)` We have, `P(Avv(BnnC)=P(A)(B)P(C)=P(A)P(B)P(C)=P(A)P(BnnC)` `impliesA and BnnC` are independent Therefore, `S_(2)` is true. Also, `P[(Ann(BuuC)]=P[(AnnB)uu(AnnB)nn(AnnC)]` `=P(AnnB)+P(AnnC)-P(AnnBnnC)` `=P(A)P(B)+P(A)P(C)-P(A)P(B)P(C)` `P(A)[P(B)+P(C)-P(B)P(C)]` `P(A)[P(B)+P(C)-P(BnnC)]` `=P(A)P(BnnC)`
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The probability of event `A` is `3//4`. The probability of event `B`, given that event `A` occurs is `1//4`. The probability of event `A`, given that
Correct Answer - C `(c )` `P(A)=(3)/(4)`, `P(B//A)=(1)/(4)` `P(A//B)=(2)/(3)` `P(B//A)=(P(BnnA))/(P(A))=(1)/(4)` (given) `:.P(BnnA)=(3)/(4)*(1)/(4)=(3)/(16)` Now `P(A//B)=(P(AnnB))/(P(B))=(2)/(3)`(given) `:.P(B)=(3)/(2)*(3)/(16)=(9)/(32)` `:. P(AuuB)=(3)/(4)+(9)/(32)-(3)/(16)=(24+9-6)/(32)=(27)/(32)` `:. P(A^(C )nnB^(C ))=1-P(AuuB)=1-(27)/(32)=(5)/(32)`
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If the trial balance does not agree the difference of the trial balance is placed in _______ account.
Correct option is (b) Suspense
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