Answer:
The values of p = 0 < p ≤ 2/3
Step-by-step explanation:
First we write the probability of 6V flashlight working representing it as a binomial mass function of a binomial random variable
P ( 6v flashlight working ) = P ( at least one 6V battery works )
= P ( 1 6v battery work ) + P ( 2 6v battery work )
= b( 2; 2, p ) + b( 1; 2, p )
write out a formula of b( 2; 2, p ) + b( 1; 2, p )
P ( 6v flashlight working ) = p^2 + 2p( 1 - p )
Next we write the probability of D flashlight working representing it as a binomial mass function of a binomial random variable
P ( D flashlight works ) = p ( at least two D batteries works )
= b( 4; 4, p ) + b(3;4, p ) + b(2; 4, p )
write out a formula of b( 4; 4, p ) + b(3;4, p ) + b(2; 4, p )
P ( D flashlight works ) = ![P^4 + 4p^3 ( 1- p ) + 6p^2 ( 1- p)^2](https://tex.z-dn.net/?f=P%5E4%20%2B%204p%5E3%20%28%201-%20p%20%29%20%2B%206p%5E2%20%28%201-%20p%29%5E2)
attached below is the remaining solution