Question

Tablet computers have became very popular and fill a gap between smart phone and PCs. The probability that the randomly selected tablet owner use their device to play games is 40%. Ten tablet owners are selected at random.

1. Find the probability that exact three of them use their device to play games.

2. What is the probability that at most three of them use their device to play games ?

3. Find the probability that between three and five (inclusive) of them use their device to play games.

Answer 1

X ~ Bin ( n , p)

Where n = 10 , p = 0.40

Binomial probability distribution is

P(X) = nCx * p^x * ( 1 - p)^n-x

1)

P(X = 3) = ¹⁰C₃ * 0.4³ * ( 1 - 0.4)⁷

= 0.2150

2)

P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= ¹⁰C₀ * 0.4⁰ * ( 1 - 0.4)¹⁰ +¹⁰C₁ * 0.4¹ * ( 1 - 0.4)⁹ +¹⁰C₂ * 0.4² * ( 1 - 0.4)⁸ +¹⁰C₃ * 0.4³ * ( 1 - 0.4)⁷

= 0.3823

3)

P(3 <= X <= 5) = P(X =3) + P(X = 4) + P(X = 5)

= ¹⁰C₃ * 0.4³ * ( 1 - 0.4)⁷ +¹⁰C₄ * 0.4⁴ * ( 1 - 0.4)⁶ +¹⁰C₅ * 0.4⁵ * ( 1 - 0.4)⁵

= 0.6665