Hybrid vigour occurs due to which event in the following ?
(A) Homozygosity
(B) Crossing over
(C) Heterozygosity
(D) Chiasma
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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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In maize, hybrid vigour is exploited by:
Correct Answer - C Hybrid vigour has been commerically exploited in different commerical crops like maize sorghum bajra tomato sugarbeet hybridistion or crossing of two unrealted individuals or parental lines leakd...
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An unbiased normal coin is tossed `n` times. Let `E_1:` event that both heads and tails are present `n` tosses. `E_2:` event that the coin shows up he
`P(E_(1))=1-["P(all heads)+P(all tails")]` `=1-[(1)/(2^(n))+(1)/(2^(n))]=1-(1)/(2^(n-1))` `P(E_(2))=P("no head)+P(exactly one head")` `=(1)/(2^(n))+^(""n)C_(1).(1)/(2^(n))=(n)/(2^(n))` `P(E_(1)nnE_(2))=("exactly one head and"(n-1)till)` `=^(""n)C_(1).(1)/(2).(1)/(2^(n-1))=(n)/(2^(n))` If `E_(1)and E_(2)` are indipendent, then `P(E_(1)nnE_(2))=P(E_(1)).P(E_(2))` `implies(n)/(2^(n))=(1-(1)/(2^(n-1)))((n+1)/(2^(n)))` `impliesn=(1-(1)/(2^(n-1)))(n+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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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.
Correct Answer - `4.0.784` Here `P(A)=0.4and P(barA)=0.6` Probability that A does not happen at all is `(0.6)^(3).` Thus, required probability is `1-(0.6)^(3)=0.784.`
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Two fair dice are rolled. Let `P(A_(i))gt0` donete the event that the sum of the faces of the dice is divisible by i. Which one of the following event
Correct Answer - A `P(A_(2))=18/36,` `P(A_(3))=12/36=1/3,` `P(A_(4))=9/36=1/4` `P(A_(5))=7/36` `P(A_(6))=6/36=1/6` Hence, `A_(3)` is most probable.
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A fair coin is flipped n times. Let E be the event "a head is obtained on the first flip" and let `F_(k)` be the event "exactly k heads are obtained".
Correct Answer - 2 Here `P(E)=1/2and P(F_(k))=""^(n)C_(k).(1)/(2^(n))` Also, `P(EnnF_(k))=p` (exactly k heads are obtained and head obtained in first filp) `=1/2""^(n-1)C_(k-1)((1)/(2))^(n-1)` Events E and `f_(k)` are independent. Therefore, `P(EnnF_(k))=P(E).P(F_(k))` `or ""^(n-1)C_(k-1)xx(1)/(2^(n))=1/2xx""^(n)C_(k)(1)/(2^(n))`...
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An unbiased normal coin is tossed n times. Let `E_(1):` event that both heads and tails are present in n tosses. `E_(2):` event that the coin shows up
Correct Answer - 3 `P(E_(1))=1-[P ("all heads)=P(all tails")]` `=1-[(1)/(2^(n))+(1)/(2^(n))]=1-(1)/(2^(n-1))` `P(E_(2))=["P(no head)+P (exactly one head)"]` `=(1)/(2^(n))+""^(n)C_(1)xx(1)/(2^(n-1))=(n)/(2^(n))` If `E_(1) and E_(2)` are independent, then `(n)/(2^(n))(1-(1)/(2^(n-1)))((n+1)/(2^(n)))` `orn =(1-(1)/(2^(n-1)))(n+1)` `or n=n+1-(n+1)/(2^(n-1))` `or n+1=2^(n-1)` `or n=3`
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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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