Question

In a city of some western country, 70% of the married persons take divorce. The probability that at least 3 among 4 persons will take divorce is​

Answer 1

Answer:

3/4 will be the probability (not sure)

Answer 2

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

We know that,

Probability of getting r success out of n independent trials with probability of success p and failure q is

$\boxed{ \bf{ \: P(r) \: = \: ^nC_r \: {p}^{r} \: {q}^{n \: - \: r}}}$

where,

• n is number of independent trials

• p is probability of success

• q is probability of failure

Now,

Given that,

➢ 70 % of the married persons took divorce

➢ It means, 30 % of the married persons didn't took divorce.

Let assume that

➢ p be the probability that married person took a divorce.

and

➢ q be the probability that married person didn't took a divorce.

➢ Now, we have to find the probability that out of 4 persons, atleast 3 will take divorce.

➢ So, we have the following parameters with us.

$\red{\rm :\longmapsto\:n = 4}$

$\red{\rm :\longmapsto\:p = \dfrac{70}{100} = \dfrac{7}{10} }$

$\red{\rm :\longmapsto\:q = \dfrac{30}{100} = \dfrac{3}{10} }$

So,

Probability of getting atleast 3 persons took divorce out of 4 is

$\rm \: = \:P(3) + P(4)$

$\rm \: = \:^4C_3 \: {\bigg[\dfrac{7}{10} \bigg]}^{3} \bigg[\dfrac{3}{10} \bigg] \: + \: ^4C_4 \: {\bigg[\dfrac{7}{10} \bigg]}^{4}$

$\rm \: = \:4 \: {\bigg[\dfrac{7}{10} \bigg]}^{3} \bigg[\dfrac{3}{10} \bigg] \: + \: 1 \times \: {\bigg[\dfrac{7}{10} \bigg]}^{4}$

$\rm \: = \: {\bigg[\dfrac{7}{10} \bigg]}^{3}\bigg[\dfrac{12}{10} + \dfrac{7}{10} \bigg]$

$\rm \: = \: {\bigg[\dfrac{7}{10} \bigg]}^{3}\bigg[\dfrac{12 + 7}{10} \bigg]$

$\rm \: = \: {\bigg[\dfrac{7}{10} \bigg]}^{3}\bigg[\dfrac{19}{10} \bigg]$

$\rm \: = \:\dfrac{6517}{10000}$

$\rm \: = \:0.6517$

Additional Information :-

• Mean of Binomial Distribution = np

• Variance of Binomial Distribution = npq

• Mean > Variance