The probability that Rajesh will score more than 90 marks in class test is 0.75. What is the probability that Rajesh will secure more than 90 marks in three out of four class tests?
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Need perfect explanation
parisakura98pari:
is its ans is 9/16?
Answers
Answered by
9
Probability of rajesh' s mark comes 90+(p) = 0.75
then , probability of Rajesh' marks 90- (q)= 0.25
n = number of total test class test = 4
r = number of test in which Rajesh got 90+ marks = 3
Here use binomial distribution ,
Probability of rajesh = B( n , P) = nCrP^r .q^( n -r)
= (4!/3!)× ( 3/4)³ × (1/4)^(4-3)
= 4 × (3/4)³× 1/4
= 27/64
then , probability of Rajesh' marks 90- (q)= 0.25
n = number of total test class test = 4
r = number of test in which Rajesh got 90+ marks = 3
Here use binomial distribution ,
Probability of rajesh = B( n , P) = nCrP^r .q^( n -r)
= (4!/3!)× ( 3/4)³ × (1/4)^(4-3)
= 4 × (3/4)³× 1/4
= 27/64
Ivam not sure , but according to data , I hope I will correct this question asked in ignou BA part 2 economics .
I think of 1 or 27/64
most probably 1 should be the answer
1 can't be possible
here only 3 events out of 4
1 means 100 percent
Perhaps it's a little difficult to understand this formula and explanation.
yeah !!! you are right sir !!! but i do this after lossing my 30 minutes .
Actually this concept related to binomials distribution . and probability but i know only basic probability . this is i think 12th last chapter
Answered by
11
Combinations of 3 tests out of total 4 tests = 4C3 = 4.
Getting 90+ marks in different tests are assumed to be independent events.
Probability that he gets <90 in a test = 1 - 3/4 = 1/4.
Probability that he gets 90+ in 3 tests and gets <90 in 4th test
= 3/4 * 3/4 * 3/4 * 1/4
= 27/256
Answer = 4C3* 27/256 = 27/64
Getting 90+ marks in different tests are assumed to be independent events.
Probability that he gets <90 in a test = 1 - 3/4 = 1/4.
Probability that he gets 90+ in 3 tests and gets <90 in 4th test
= 3/4 * 3/4 * 3/4 * 1/4
= 27/256
Answer = 4C3* 27/256 = 27/64
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thank you sir
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