Question

A die is thrown 6 times. If “getting an odd number" is a "success", what is the probability
of getting at least one success.

Answer 1

Answer:

1/2

Step-by-step explanation:

it is a perfect answer

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given that,

A die is rolled 8 times and getting an even number is considered as success denoted by event A.

So, number of independent trials, n = 8.

Let assume that

p denotes the probability of success i.e getting an even number.

and

q denotes the failure i.e. not getting an even number.

The possible outcomes of even number when a die is rolled is { 2, 4, 6 }

So,

$\rm :\longmapsto\:p = \dfrac{3}{6} = \dfrac{1}{2}$

and

$\rm :\longmapsto\:q = 1 - p = 1 - \dfrac{1}{2} = \dfrac{1}{2}$

Now, We know In Binomial Distribution,

In a random experiment, consisting of n number of independent trials, the probability of getting r successes is

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: P(r) \: = \: ^nC_r \: {p}^{r} \: {q}^{n - r} \: }}}$

a) The probability that event A is success 5 times.

$\red{\rm :\longmapsto\:P(5)}$

$\rm \: = \: ^8C_5 \: {\bigg[\dfrac{1}{2} \bigg]}^{5} {\bigg[\dfrac{1}{2} \bigg]}^{3}$

$\rm \: = \: \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \dfrac{1}{32} \times \dfrac{1}{8}$

$\rm \: = \: \dfrac{7}{32}$

b) The probability of not getting A at all.

$\red{\rm :\longmapsto\:P(0)}$

$\rm \: = \: ^8C_0 \: {\bigg[\dfrac{1}{2} \bigg]}^{8} {\bigg[\dfrac{1}{2} \bigg]}^{0}$

$\rm \: = \: \dfrac{1}{256}$

c) The probability of getting A at least one time.

$\red{\rm :\longmapsto\: \displaystyle \sum^8_{r = 1} \: P(r)}$

$\rm \: = \: P(1) + P(2) + P(3) + - - + P(8)$

$\rm \: = \: 1 - P(0)$

$\rm \: = \: 1 - \dfrac{1}{256}$

$\rm \: = \: \dfrac{256 - 1}{256}$

$\rm \: = \: \dfrac{255}{256}$

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Additional Information :-

1. Mean of Binomial Distribution = np

2. Variance of Binomial Distribution = npq

3. Mean > Variance