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A basketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. She misses the first shot 40% of the time. When she misses the first shot, she misses the second shot 5% of the time. What is the probability of missing both free-throw shots?

Question

Help i medal and fan thanks!!!!!!!!
A basketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. She misses the first shot 40% of the time. When she misses the first shot, she misses the second shot 5% of the time. What is the probability of missing both free-throw shots?

anonymous · Answer

40%=0.40
5%=0.05
You just multiply them together

itsmichelle29 · Answer

so 2%

itsmichelle29 · Answer

what the formula for that

anonymous · Answer

It is used when 2 or more independent events are occurring. The formula would be percentage of probability 1 x percentage of probability 2 
or 

percentage of probability 1 x percentage of probability 2 x percentage of probability 3 
if there are 3 independent event and you want to find the probability of it

itsmichelle29 · Answer

so is it 2% or 45%

anonymous · Answer

You were right its 2%. 
45% is too off. Think about it.

itsmichelle29 · Answer

IDK i think its 45

itsmichelle29 · Answer

P(A U B)

anonymous · Answer

No, you multiply 0.05 x 0.40 as I stated earlier which equals to 2. Why would she have a 45% chance to miss BOTH free throws when she has a 5% chance to miss it after missing her first one? Reading the question 5 more time to fully understand the question.

itsmichelle29 · Answer

Okay thanks im going with what you say thanks ... im wrong

anonymous · Answer

Yeah, no problem. Feel free to tell me the results.