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Sagot :
The probability of getting exactly 4 heads is 0.26
A loaded coin has 0.48 probability of coming up heads
0.52 probability of coming up tails.
coin is flipped 7 times
let's assume n = 7.
The seven flips of the coin are separate from one another, so what happens on one flip is unaffected by the outcomes of the other tosses.
When calculating the likelihood that two events will occur simultaneously, we must multiply the probabilities of events 1 and 2, presuming that event 1 occurred.
Tosses never affect one another, so the "assuming event 1 happened" part is no longer necessary because it has no effect on the probability. As a result, we just need to multiply the chances for each toss. This supposition is essential. We would have needed to know exactly how tosses are dependent on one another if they were interdependent.
Typically, we would have [tex](n! /(r!)*(n - r)!)*p^r*(1-p)^(n-r)[/tex] is the probability of getting r heads out of n, without taking into account the order.
probability of getting exactly 4 heads r = 4.
out of 7 flipping n =7.
7!/(4!*3!) = 35
(0.48)^4 = 0.05308416
(0.52)^3 = 0.140608
35*0.05308416*0.140608 = 0.26124201492 = 0.26
so the probability of getting exactly 4 heads is 0.26.
To know more about probability here,
https://brainly.com/question/23800890
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