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# A coin is tossed 5 times. Find the probability of getting (1) at least 4 heads and (2) at most 4 heads.  Hint: First we have to know the meaning of probability, it is the ratio of total number of desired outcomes to total number of possible outcomes and we have to see various desired outcomes to us and total no of possible outcomes and then we will get probability.

Complete step-by-step solution -
Given that coin is tossed up 5 times
So the total number of possible outcomes is ${{2}^{^{5}}}$
Probability of getting at least 4 heads:
Probability of getting at least 4 heads is given by probability of getting 4 heads and getting 5 heads
No of ways of getting 4 heads is $\left\{ HHHHT,THHHH,HTHHH,HHTHH,HHHTH \right\}$
So, number of ways of getting 4 heads is 5
No of ways of getting 5 heads is $\left\{ HHHHH \right\}$
So, no ways of getting 5 heads is one
Probability of getting at least 5 heads is
$\dfrac{5+1}{32}=\dfrac{3}{16}$
Probability of getting at most 4 heads:
Probability of getting utmost 4 heads is total minus probability of getting 5 heads
No. of ways of getting 5 heads is 1.
So, the probability of getting 5 heads is $\dfrac{1}{32}$
Probability of getting utmost 4 heads is total probability minus probability of getting 5 heads
$=1-\dfrac{1}{32}$
$=\dfrac{31}{32}$

Note: The value of probability always lies between 0 and 1 but it should not be negative and it should not be greater than 1 because the total number of possible outcomes is always greater than no of desired outcomes. In the second case there is only one outcome which has 5 heads so for at most 4 heads we will have only 31 outcomes. Sometimes students make mistakes in finding all the 31 outcomes and it becomes time taking.
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