A coin is tossed 3 times. at least 1 head is obtained. Determine each probability...?(adsbygoogle = window.adsbygoogle || []).push({});

1) exactly 1 head is obtained

2) exactly 2 heads are obtained

3) exactly 3 heads are obtained.

------------------------------------------------------------

for 1) brute force indicates that there 7 possible combinations:

HHH, HHT, HTH, HTT, THH, THT, TTH (because at least one heads is obtained).

Out of these we see that there are 3 occasions where there is exactly one heads.

If we let S be the sample space of all possable outcomes, and let E be the event of having exactly one heads, then a basic rule of probability indicates that the probability P(E)=|E|/|S|=3/7.

However, using another method:

fix one coin throw's result as heads. Then the probability of the other two throws being tails is (1/2)(1/2)=1/4.

Since we can fix heads 3 times, this indicates that the probability is 3(1/4)=3/4.

Can anyone spot where the flaw is in either of these attempts at a solution?

thanks

(this isn't a hw question btw)

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Tricky probability question

**Physics Forums | Science Articles, Homework Help, Discussion**