Yup, rotten and not rotten => 2 outcomes. Sometimes when the probability is not 50% it throws people off.
I have this question that I think is binomial probability.
Bananas that are almost ripe to pick carry the chance of being 15% rotten. What would be the probability of having at most 2 rotten bananas packed in a box of 12?
Yup, rotten and not rotten => 2 outcomes. Sometimes when the probability is not 50% it throws people off.
Stop cowardice, ban guns!
Hi Chris,
did you found the sollution?
if so, could you share it with us?
thanks
You just need to plug the numbers into the binomial distribution, which can be done using a table or software, so what is the probability of 0, 1, and 2 rotten tomatoes?
Stop cowardice, ban guns!
Probability of picking a rotten bananais 0.15. The probability remains constant at 0.15. The fact that you have picked a banana does not affect the probablity when picking the next banana. You therefore have independent trials. There are only two outcomes (rotten or not rotten).
Three conditions for binomial distribution are therefore satisfied 1) Only two possible outcomes 2) Independent events 3) probability remains constant.
Problem says you can have 0 or 1 or 2 and the condition (not more than 2) is satisfied. Probability that x=0 + probability that x=1 + probability that x=2 gives the answer. If your calculations do not add up to 0.736 then I have made a mistake or you have made a mistake
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