Question

If hens of a certain breed lay eggs on 5days a week on an average. Find how many days during a session of 100 days a poultry keeps with 5 hands of this breed 'will expect to receive at least 4 eggs

Answer 1

Answer:

p=5/7= probability of one hen laying an egg on a given day

q= 2/7 = probability of one hen not laying egg

probability of laying atleast 4 eggs by 5 hens = 5C4(5/7)^4(2/7)^1 + 5C5(5/7)^5(2/7)^0

=.5577

the expected no of days during a session of 100 days during which atleast 4 eggs are laid by 5 hens are...100×p = 100×.5578=56 days

Answer 2

Answer:

56 Days

Step-by-step explanation:

Number of Eggs Required = 100 * 4 = 400 egss

Number of eggs by 1 hen in 1 week = 5  eggs

Number of eggs by 5 hen in 1 week = 25  eggs

Number of week required to get 400 eggs = 400/25

=> Number of week required to get 400 eggs = 8 Weeks

Number of Days in 1 Week  = 8 * 7 = 56 Days