Question

Ten little monkeys were jumping on a bed. There is a 35% chance that one will fall off and bump his head. In the bedroom next door, five kangaroos were jumping on a bed. Being more adept at jumping, there is only a 20% that a kangaroo will fall off the bed.
1. What are the chances that a monkey and a kangaroo will fall off the bed?
2. What are the chances that a monkey will not fall off the bed?
3. If the monkeys enjoy this activity every night for an entire week, what are the chances that a monkey falls off the bed every one of the seven nights?
4. What are the chances that if the monkeys jump every day for a week that at least one will fall off of the bed?
5. What are the chances that the kangaroos can get away with jumping on the bed for 4 straight nights until they finally have someone fall off the bed on the night?
6. What are the chances that the kangaroos can jump two nights in a row with no one falling off the bed?
7. The monkeys manage to go a whole week without someone bumping their head. One of the kangaroos insists that they are due for an injury. Another says they must be getting better at their jumping skills. Do you think they're due for a crash?

Answer 1

Answer:

0.07 ; 0.65 ; 0.00064339296875 ; 3.02 * 10^-12

Step-by-step explanation:

Given :

Number of monkeys, n1 = 10

Probability of fall, off ; p(m) = 35% = 0.35

Number of kangaroos, n2 = 5

Probability of fall, off : p(k) = 20% = 0.2

1.) A monkey and kangaroo falls off;

p(m) * p(k) = 0.35 * 0.2 = 0.07

2.) probability that monkey will not fall off;

1 - p(m) = 1 - 0.35 = 0.65

3.) probability that a monkey falls off every one of seven nights :

P(m) = 0.35 ; for 7 nights ;

0.35 *0.35 * 0.35 * 0.35 * 0.35 * 0.35 * 0.35 = 0.35^7 = 0.00064339296875

4.)

P(x ≥ 1) = p(1) + p(2) +... + p(10)

Using binomial probability calculator :

n = 7*10 = 70 ; p = 0.35

P(x ≥ 1) = 70C1 * 0.35^1 * 0.65^69 = 3.02 * 10^-12