Question

Find the probability of each event. A six-sided die is rolled seven times. What is the probability that the die will show an even number at most five times?

Answer 1

Answer:

$\dfrac{15}{16}$

Step-by-step explanation:

When a six sided die is rolled, the possible outcomes can be:

{1, 2, 3, 4, 5, 6}

Even numbers are {2, 4, 6}

Odd Numbers are {1, 3, 5}

Probability of even numbers:

$\dfrac{\text{Favorable cases}}{\text{Total cases }} = \dfrac{3}{6} = \dfrac{1}{2}$

This is binomial distribution.

where probability of even numbers,  $p =\frac{1}{2}$

Probability of not getting even numbers (Getting odd numbers) $q =\frac{1}{2}$

Probability of getting r successes out of n trials:

$P(r) = _nC_r\times p^r q^{n-r}$

Probability of getting even numbers at most 5 times out of 7 is given as:

P(0) + P(1) +P(2) + P(3) +P(4) + P(5)

$\Rightarrow _7C_0\times \frac{1}{2}^0 \frac{1}{2}^{7}+_7C_1\times \frac{1}{2}^1 \frac{1}{2}^{6}+_7C_2\times \frac{1}{2}^2 \frac{1}{2}^{5}+_7C_3\times \frac{1}{2}^3 \frac{1}{2}^{4}+_7C_4\times \frac{1}{2}^4 \frac{1}{2}^{3}+_7C_5\times \frac{1}{2}^5 \frac{1}{2}^{2}$

$\Rightarrow (\dfrac{1}{2})^7 (_7C_0+_7C_1+_7C_2+_7C_3+_7C_4+_7C_5)$

$\Rightarrow (\dfrac{1}{2})^7 (1+7+\dfrac{7 \times 6}{2}+\dfrac{7 \times 6 \times 5}{3\times 2}+\dfrac{7 \times 6 \times 5}{3\times 2}+\dfrac{7 \times 6}{2})\\\Rightarrow \dfrac{120}{128} \\\Rightarrow \dfrac{15}{16}$