You have a flashlight that takes 2 working batteries. You have 8 batteries but only 4 of them work. What is the fewest number of pairs you need to test to guarantee you can get the flashlight on?
Answers
Three pairs, because there are only 4 defective batteries. If we take 3 pairs, we can assure that atleast one pair will be properly functional out of the other two (considering the worst case).
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7 pairs must you try in order to be certain that you can put the torch on when four of the eight cells you have are functional.
Given that,
You have a flashlight that requires two good batteries to operate. Four of the eight cells you have are functional.
We have to find how many pairs must you try in order to be certain that you can put the torch on.
We know that,
Total batteries = 8.
Out of the 8 batteries, 4 are working.
We need 2 good batteries to run.
Dividing the 8 batteries in a group of 3.
8 = 3 + 3 + 2
Group 1 = 3 batteries,
Group 2 = 3 batteries,
Group 3 = 2 batteries.
Group 1: B₁, B₂, B₃.
Possible combinations → B₁B₂ , B₂B₃ or B₁B₃.
Total combinations = 3
Group 2: B₄ B₅ B₆.
Possible combinations → B₄B₅ , B₅B₆ or B₄B₆.
Total combinations = 3.
Group 3: B₇ B₈.
Possible combinations → B₇B₈.
Total combinations = 1.
Adding all the combinations = 3 (Group 1) + 3 (Group 2) + 1 (Group 3) =7.
Therefore, minimum number of trails needed = 7.
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