A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots.

[DivyaD]

A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

a) $$\frac{1}{6}$$
b) $$\frac{625}{1296}$$
c) $$\frac{649}{1296}$$
c) $$\frac{671}{1296}$$
d) $$\frac{2}{3}$$

[Brent@GMATPrepNow]

A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

a) $$\frac{1}{6}$$
b) $$\frac{625}{1296}$$
c) $$\frac{649}{1296}$$
c) $$\frac{671}{1296}$$
d) $$\frac{2}{3}$$

We want P(select at least one 6)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least one 6) = 1 - P(not getting at least one 6)
What does it mean to not get at least one 6? It means getting zero 6's.
So, we can write: P(getting at least one 6) = 1 - P(getting zero 6's)

P(getting zero 6's)
P(getting zero 6's) = P(no 6 on 1st roll AND no 6 on 2nd roll AND no 6 on 3rd roll AND no 6 on 4th roll)
= P(no 6 on 1st roll) x P(no 6 on 2nd roll) x P(no 6 on 3rd roll) x P(no 6 on 4th roll)
= 5/6 x 5/6 x 5/6 x 5/6
= 625/1296

P(getting at least one 6) = 1 - 625/1296
= 671/1296

Answer: D

Cheers.
Brent

[fskilnik@GMATH]

A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

$$ a) \frac{1}{6}\,\,\,\,\,\,\, b) \frac{625}{1296}\,\,\,\,\,\,\,c) \frac{649}{1296}\,\,\,\,\,\,\,d) \frac{671}{1296}\,\,\,\,\,\,\,e) \frac{2}{3}$$

$$? = 1 - P\left( {\underbrace {{\rm{no}}\,\,{\rm{6}}\,\,{\rm{in}}\,\,{\rm{all}}\,\,{\rm{4}}\,\,{\rm{times}}}_{{\rm{unfavorable}}}} \right)$$
$$\left. \matrix{
{\rm{Total}}\,\,:\,\,\,{6^4}\,\,{\rm{equiprobable}}\,\,{\rm{outcomes}}\,\,\, \hfill \cr
{\rm{unfav}}\,\,{\rm{:}}\,\,{5^4}\,\,{\rm{of}}\,\,{\rm{them}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,P\left( {{\rm{unfavorable}}} \right) = {\left( {{5 \over 6}} \right)^4}$$
$$? = {{{6^4} - {5^4}} \over {{6^4}}} = {{\left( {{6^2} + {5^2}} \right)\left( {6 + 5} \right)\left( {6 - 5} \right)} \over {{6^4}}} = {{61 \cdot 11} \over {{6^4}}} = {{610 + 61} \over {{6^4}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.