Question

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing lights in all directions when no other cars are visible. What is the probability that, of 15 randomly chosen drivers coming to an intersection under these conditions

a) At most 6 will come to a complete stop?

b) Exactly 6 will come to a complete stop?

c) At least 6 will come to a complete stop?

d) How many of next 20 drivers do you expect to come to complete stop?

Please show work when completely this. I am really confused.

Answer 1

X ~ Binomial (n,p)

Where n = 15 , p = 0.20

Binomial probability distribution is

P(X) = ⁿC_x p^x ( 1- p)^n-x

a )

P( at most 6) = P( X <= 6)

= P( X = 0) + P( X = 1)+P( X = 2)+P( X = 3)+P( X = 4)+P( X = 5)+P( X = 6)

= ¹⁵C₀ 0.20⁰ 0.80¹⁵ +¹⁵C₁ 0.20¹ 0.80¹⁴ +¹⁵C₂ 0.20² 0.80¹³ +¹⁵C₃ 0.20³ 0.80¹² +

¹⁵C₄ 0.20⁴ 0.80¹¹ +¹⁵C₅ 0.20⁵ 0.80¹⁰ +¹⁵C₆ 0.20⁶ 0.80⁹

= 0.9819

b)

P( X = 6) = ¹⁵C₆ 0.20⁶ 0.80⁹

= 0.0430

c)

P(at least 6) = P( X >= 6)

= 1 - P( X <= 5)

= 1 - [ ¹⁵C₀ 0.20⁰ 0.80¹⁵ +¹⁵C₁ 0.20¹ 0.80¹⁴ +¹⁵C₂ 0.20² 0.80¹³ +¹⁵C₃ 0.20³ 0.80¹² +

¹⁵C₄ 0.20⁴ 0.80¹¹ +¹⁵C₅ 0.20⁵ 0.80¹⁰ ]

= 1 - 0.9389

= 0.0611

d)

Here number of drivers n = 20

E(X) = n * p

= 20 * 0.20

= 4

Of the next 20 drivers number of drivers we expect to come to complete stop = 4