Question:
i want to ask taht what is the shortcut method of the question like when a coi is tossed 8times getting a head is the succes . what is the probability that atleast 2 heads wiil occur
Answer:
Let x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)
The general form for probability of random variable x is
P(X = x) = nCx * px * qn-x
Now, in the question, we want at least two heads
Now, p =q = 1/2
So, P(X ≥ 2) = 8C2 * (1/2)2 * (1/2)8-2
=> P(X ≥ 2) = 8C2 * (1/2)2 * (1/2)6
=> 1 - P(X < 2) = 8C0 * (1/2)0 * (1/2)8 + 8C1 * (1/2)1 * (1/2)8-1
=> 1 - P(X < 2) = (1/2)8 + 8 * (1/2)1 * (1/2)7
=> 1 - P(X < 2) = 1/256 + 8 * (1/2)8
=> 1 - P(X < 2) = 1/256 + 8/256
=> 1 - P(X < 2) = 9/256
=> P(X < 2) = 1 - 9/256
=> P(X < 2) = (256 - 9)/256
=> P(X < 2) = 247/256
In this way, we can solve this type of problem.
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