Questions & Answers

Question

Answers

A. $\dfrac{1}{{16}}$

B. $\dfrac{2}{{16}}$

C. $\dfrac{{14}}{{16}}$

D. $\dfrac{{15}}{{16}}$

Answer
Verified

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.

$\begin{array}{c}\Pr {\rm{obability}}\;{\rm{that}}\;{\rm{at}}\;{\rm{least}}\;{\rm{one}}\;{\rm{head}}\;{\rm{occurs}} = {P_1} + {P_2} + {P_3} + {P_4}\\ = \dfrac{1}{{16}} + \dfrac{4}{{16}} + \dfrac{6}{{16}} + \dfrac{4}{{16}}\\ = \dfrac{{15}}{{16}}\end{array}$

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