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( B )dot`
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) + P(B) - 0.2`
` or P(A) + P(B) = 0.8`
or `1 - P(bar(A)) + 1- P(bar(B))= 0.8`
or `P(barA)+P(bar(B)) = 1.2`
Correct option is: D)\(\frac{3}{4}\)
When two coins tossed simultaneously then possible outcomes are (HH, HT, TH, TT)
\(\therefore\) Total possible outcomes = 4.
Favourable outcomes to event of getting at least one head are (HH,...
Correct option is: D)\(\frac{7}{8}\)
When three coins are tossed, then possible outcomes are
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\therefore\) Total possible outcomes = n(s) = 8
Outcomes which favours the event that at...
Correct option is: A) \(\frac{7}{8}\)
When three coins are tossed simultaneously once.
Then total possible outcomes are
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\therefore\) Total possible outcomes is n (S) = 8
Favourable outcomes to the...
We know that, `AcupB` denotes the occurance of atleast one of A and B and `AcapB` denotes the occurance of both A and B, simuletanously.
Thus, `P(AcupB)=0.6and P(AcapB)=0.3`
Also, `P(AcupB)=P(A)+P(B)_P(AcapB)`...
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...
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.`
Correct Answer - A::D
Let `P(E)=eand P(F)=f`
`P(EuuF)-P(EnnF)=11/25`
`impliese+f-2ef=11/25" "(1)`
`P(barEnnbarF)=2/25`
`implies(1-e)(1-f)=2/25" "(2)`
From (1) and (2),
`ef=12/25and e+f=7/5`
Solving, we get
`e=4/5,f=3/5or e=3/5,f=4/5`
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