A baised coin is tossed thrice with the probability of head 60 %. find the probability of at least 2 tails come
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This is a binomial distribution.
Bin (n, p) = ⁿₓPˣ (1 - P)ⁿ⁻ˣ
n = number of tosses
p = probability of getting a head
Doing the substitution we have :
Bin (3,0.6) = ³x(0.6)ˣ (1 - 0.6)³⁻ˣ
x = 0,1, 2, 3
The probability that a tail is obtained at least twice.
= P( x = 0) or P(x = 1)
Doing the substitution we have :
³₀(0.6)⁰ (0.4)³⁻⁰ + ³₁(0.6) (0.4)³⁻¹
= 3!/(3 - 0)! 0! × 0.6 ⁰× 0.4³ = 0.064
= 3!/(3 - 1)! 1! × 0.6 × 0.4² = 0.288
Adding the two :
0.064 + 0.288 = 0.352
The answer is thus : 0.352
L
Bin (n, p) = ⁿₓPˣ (1 - P)ⁿ⁻ˣ
n = number of tosses
p = probability of getting a head
Doing the substitution we have :
Bin (3,0.6) = ³x(0.6)ˣ (1 - 0.6)³⁻ˣ
x = 0,1, 2, 3
The probability that a tail is obtained at least twice.
= P( x = 0) or P(x = 1)
Doing the substitution we have :
³₀(0.6)⁰ (0.4)³⁻⁰ + ³₁(0.6) (0.4)³⁻¹
= 3!/(3 - 0)! 0! × 0.6 ⁰× 0.4³ = 0.064
= 3!/(3 - 1)! 1! × 0.6 × 0.4² = 0.288
Adding the two :
0.064 + 0.288 = 0.352
The answer is thus : 0.352
L
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