Question

In a family the husband tells a lie in 30% cases and the wife in 35% cases. Find the probability that both state the same fact.

Answer 1

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$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
In a family,
Husband tells a lie in 30% cases
And his wife in 35% cases.

Probability that both state the same fact = ?
(or) Probability that Husband and Wife contradict each other = ?

$\underline{\large{\textbf{Step-1:}}}$
Assume two Events and find probability of the events.

Let H be the Event that Husband tells the lie
and W be the Event that Wife tells the lie

$P(H) = \dfrac{30}{100} = \dfrac{3}{10}$ --------(1)

$P(W) = \dfrac{35}{100} = \dfrac{7}{20}$---------(2)

$\underline{\large{\textsf{Step-2:}}}$
Note the Complementary Events

Opposite action:
Telling the lie $\times$ Telling the truth

Let $\overline{H}$ be the Event that Husband tells the truth

and $\overline{W}$ be the Event that Wife tells the truth.

$\underline{\large{\textsf{Step-3:}}}$
Find the Value of Complementary Events

We have,
Sum of Probabilities = 1

$\implies P(H) + P(\overline{H}) = 1$

$\implies P(\overline{H}) = 1 - P(H)$

$\implies P(\overline{H}) = 1 - \dfrac{30}{100}$

$\implies P(\overline{H}) = \dfrac{100-30}{100}$

$\implies P(\overline{H}) = \dfrac{70}{100}$

$\implies P(\overline{H}) = \dfrac{7}{10}$ ---------(3)

Also,

$\implies P(W) + P(\overline{W}) = 1$

$\implies P(\overline{W}) = 1 - P(W)$

$\implies P(\overline{W}) = 1 - \dfrac{7}{20}$

$\implies P(\overline{W}) = \dfrac{20-7}{20}$

$\implies P(\overline{W}) = \dfrac{13}{20}$---------(4)

$\underline{\large{\textsf{Step-4:}}}$
Find the Probability that Husband and Wife contradict each other.

P(H and W contradict each other)

= P(H tells the lie and W tells the truth) or P(H tells the truth and W tells tells the lie)

= P(H and $\overline{W}$) or P($\overline{H}$ and W)

Here,
We assume that,
case-(1) If H tells the truth, W tells the lie
case-(2)If H tells the lie, W tells the truth.

Here,
Only one of the case can Occur,
i.e., they are Mutually Exclusive.

We have,
If A and B are Mutually Exclusive(or Disjoint) events, Probability of happening of two disjoint events A or B is
$\bigstar$ P(A or B) = P(A) + P(B)

Thus,

P(H and W contradict each other)
= P(H and $\overline{W}$) + P($\overline{H}$ and W) -----------(5)

Here,
H and $\overline{W}$ are said to be Independent Events
Similarly, W and $\overline{H}$ are said to be Independent Events

Since,
Occurrence of One Event Doesn't affect the Occurrence of Other.

We have,
If A and B are Independent Events, Probability of happening of two events A and B
$\bigstar$ P(A and B) = P(A)$\times$P(B)

Therefore,
P(H and $\overline{W}$) = P(H) $\times$ P($\overline{W}$) -------(6)

P($\overline{H}$ and W) = P($\overline{H}$) $\times$ P(W) ---------(7)

Substituting (6)&(7) in (5)

$\implies$
P(H and W contradict each other)
= [P(H) $\times$ P($\overline{W}$)] + [P($\overline{H}$) $\times$ P(W)]

Substituting Values (1), (2), (3), (4)

P(H and W contradict each other)

= $[(\dfrac{3}{10}) \times (\dfrac{13}{20})]+[(\dfrac{7}{10}) \times (\dfrac{7}{20})]$

= $\dfrac{39}{200} + \dfrac{49}{200}$

= $\dfrac{39+49}{200}$

= $\dfrac{88}{200}$

= $\dfrac{11}{25}$

Expressing in Percentage

= $(\dfrac{11}{25} \times 100)\%$

= $(11\times 4)\%$

P(H and W contradict each other)
= $44\%$

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