Let p be a probability function on set S, where S = (a1, a2, a3, a4). Find P(a1) if P(a2) = \(\frac{1}{3}\), p(a3) = \(\frac{1}{6}\) and p(a4) = \(\frac{1}{5}\)
\(\frac{7}{3}\)
\(\frac{2}{3}\)
\(\frac{1}{3}\)
\(\frac{3}{10}\)
Explanation
No explanation available
Video Explanation
No video available
Post your Contribution
Discussions (10)
Since we know that the sum of probability is 1
S = P(a1) + P(a2) + P(a3) + P(a4)
1 = P(a1) + 1/3 + 1/6 +1/5
1 = P(a1) + 7/10
P(a1) = 1 - 7/10
P(a1) = 3/10
3/10 is option D
i think it should be like this
required outcome=4
pr(a)(4) =1/3*1/6*1/5
3/1*1/6*1/5
where :3 into 6 is 2 making it
1/1*1/2*1/5
=4- 1/10
which =3/10
get to think of it now
alright alright my sweeties, here is the solution, it may look a bit complex at first but all you gotta do i not be afraid and use your mathematical Talent/Creativity, they want us to find the probability of A no-big deal... then, they gave us every other probability for B, C, D... except for the one they want us to find... I need not say no more you should have figured out where I'm headed, and if you haven't... relax you will, just get a pen and paper.
alright alright my sweeties, here is the solution, it may look a bit complex at first but all you gotta do i not be afraid and use your mathematical Talent/Creativity, they want us to find the probability of A, no-big deal... then, they gave us every other probability for B, C and D. that is all except for the one they want us to find... I dont think i need to say more you should have figured out where I'm headed, and if you haven't... relax you will, just get a pen and paper.
