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# A coin is tossed up 4 times. The probability that at least one head turn up isA. $\dfrac{1}{{16}}$B. $\dfrac{2}{{16}}$C. $\dfrac{{14}}{{16}}$D. $\dfrac{{15}}{{16}}$  Complete step by step solution:
When we tossed a coin four times the outcomes can be as
[{HHHH}, {HHHT}, {HHTH}, {HTHH}, {THHH}, {HHTT}, {HTHT}, {THHT}, {THTH}, {TTHH}, {HTTH}, {HTTT}, {THTT}, {TTHT}, {TTTH}, {TTTT}]
From the outcomes it can be seen that
Events that all four are head = {HHHH}
Number of outcomes when four heads occurs = 1
Let ${P_1}$ be the probability that all four are heads.
It is known that the probability is the number of favorable outcomes to the total number of outcomes.

Thus,$\begin{array}{c}{P_1} = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{outcomes}}\;{\rm{when}}\;{\rm{four}}\;{\rm{are}}\;{\rm{heads}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \dfrac{1}{{16}}\end{array}$
Outcomes of the experiment when three heads occurs = {HHHT}, {HHTH}, {HTHH}, {THHH}
Number of outcomes that three heads occurs = 4

Let ${P_2}$ be the probability that three heads occur.
Thus,$\begin{array}{c}{P_2} = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{outcomes}}\;{\rm{that}}\;{\rm{three}}\;{\rm{heads}}\;{\rm{occurs}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \dfrac{4}{{16}}\end{array}$
Outcomes of the experiment when two heads occurs = {HHTT}, {HTHT}, {THHT}, {THTH}, {TTHH}, {HTTH}

Number of outcomes that two heads occurs = 6
Let ${P_3}$ be the probability that three heads occur.
$\begin{array}{c}{P_3} = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{outcomes}}\;{\rm{that}}\;{\rm{two}}\;{\rm{heads}}\;{\rm{occurs}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \dfrac{6}{{16}}\end{array}$
Only one head occurs = {HTTT}, {THTT}, {TTHT}, {TTTH}

Number of outcomes that one heads occurs = 4
Let ${P_4}$ be the probability that three heads occur.
$\begin{array}{c}{P_4} = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{outcomes}}\;{\rm{that}}\;{\rm{one}}\;{\rm{heads}}\;{\rm{occurs}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \dfrac{4}{{16}}\end{array}$

Now from this information we can find the probability for the occurrence of at least one head by adding all these probabilities.
$\begin{array}{c}\Pr {\rm{obability}}\;{\rm{that}}\;{\rm{at}}\;{\rm{least}}\;{\rm{one}}\;{\rm{head}}\;{\rm{occurs}} = {P_1} + {P_2} + {P_3} + {P_4}\\ = \dfrac{1}{{16}} + \dfrac{4}{{16}} + \dfrac{6}{{16}} + \dfrac{4}{{16}}\\ = \dfrac{{15}}{{16}}\end{array}$

Hence, the correct answer is D.

Note: Probability explains the chances of occurrence of an event. Here we have to determine the probability of occurrence of at least one head in four tosses. The probability of occurrence of one head, two heads, three heads or four heads can be calculated from the given information. Thus, by adding all the probabilities, the probability of occurrence of at least one head can be calculated easily.
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