A friend recently planned a camping trip. He had two flashlights, one that required a single 6-V battery and another that used two size-D batteries. He had previously packed two 6-V and four size-D batteries in his camper. Suppose that theprobability than any particular battery works is p and that batteries work or fail independently of one another. Our friend wants to take just one flashlight. For what values of p should he take the6-V flashlight?
Answers
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 ) =
attached below is the remaining solution
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