Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If a certain coin is flipped, it has probability 1/2 of landing on heads and probability 1/2 of landing on tails. If the coin is flipped 4 times, what is the probability that it will land on tails at least once?

A. 3/8
B. 1/4
C. 1/2
D. 5/16
E. 15/16

$$? = P\left( {{\text{at}}\,\,{\text{least}}\,\,{\text{1}}\,\,{\text{tail}}\,\,{\text{in}}\,\,{\text{4}}\,\,{\text{landings}}} \right)$$
$$?\,\,\, = \,\,\,1 - P\left( {{\text{only}}\,\,{\text{heads}}} \right)\,\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,1 - {\left( {\frac{1}{2}} \right)^4} = \,\,\,\frac{{15}}{{16}}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left( {\text{E}} \right)$$
$$\left( * \right)\,\,\,P\left( {{\text{head}}\,\,{\text{and}}\,\,{\text{head}}\,\,{\text{and}}\,\,{\text{head}}\,\,{\text{and}}\,\,{\text{head}}} \right)\,\,\,\mathop = \limits^{{\text{independency}}} \,\,\,P\left( {{\text{head}}} \right) \cdot P\left( {{\text{head}}} \right) \cdot P\left( {{\text{head}}} \right) \cdot P\left( {{\text{head}}} \right)$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>

The complementary case is the event with all heads, which has probability (1/2)^4.
The probability of at least one tail is 1 - (1/2)^4 = 1 - 1/16 = 15/16.

Therefore, the answer is E.
Answer: E

Probability of at least one formular = 1 - probability of opposite event

$$\Pr obability\ of\ head\ only=\frac{1}{2}$$
$$\Pr obability\ of\ tail\ only=\frac{1}{2}$$

If coin is flipped 4 times ; Probability of tails only
$$=\left(\frac{1}{2}\right)^4$$
$$=\left(\frac{1^4}{2^4}\right)^{ }$$
$$=\left(\frac{1^4}{2^4}\right)=\frac{1}{16}$$

Probability of at least one tail
$$=1-\frac{1}{16}$$
$$=\frac{1}{1}-\frac{1}{16}$$
$$=\frac{1}{1}-\frac{1}{16}$$
$$=\frac{\left(16-1\right)}{16}$$
$$=\frac{\left(15\right)}{16}$$
$$answer\ is\ Option\ E$$