A coin is tossed 3 times. What is the probability that the number of tails obtained will be between 1 and 3 inclusive?
1 answer:
Answer:
Step-by-step explanation:
Given : When a coin is tossed twice , the total outcomes will become
HHH , HHT , HTH , THH , TTH , THT , HTT ,TTT
Total outcomes = 8
Favorable outcomes to obtain 1 and 3 inclusive = HHT , HTH , THH , TTH , THT , HTT ,TTT
Number of favorable outcomes = 7
Probability that the number of tails obtained will be between 1 and 3 inclusive =
Hence, the probability that the number of tails obtained will be between 1 and 3 inclusive
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Answer:
Step-by-step explanation:
We know i×i=i^2 which has value of -1.
We will multiply numerator and denominator by i so the denominator will no longer contain imaginary part(s).
Also multiplying by i/i does not change the value of the fraction because i/i=1.
Numerator × i gives (-5+i)i=-5i+i^2=-5i-1
=-1-5i.
Denominator × i gives (2i)i=2i^2=-2.
So the simplified version of this fraction given is:
The last simplification come from me multiplying fraction by -1/-1.