
A speaks truth 2 out of 3 times, and B 4 times out of 5; they agree in the assertion that from a bag containing 6 balls of different colors a red ball has been drawn. Find the probability that the statement is true.
Answer
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Hint: The ball is drawn at random with equal probability from the bag and A and B both see the color and decide independently whether or not to tell the truth. If they lie, they choose another of the five colors, again independently and with equal probability.
Complete step-by-step answer:
Let the probability of the event such that A speaks the truth is P(A) and the probability of the event such that B speaks the truth is P(B).
Similarly, the probability of the event such that A does not speak the truth is P(A’) and the probability of the event such that B does not speak the truth is P(B’).
So here we are given,
\[P(A)=\dfrac{2}{3}\],\[P(B)=\dfrac{4}{5}\],\[P(A')=1-P(A)=\dfrac{1}{3}\],\[P(B')=1-P(B)=\dfrac{1}{5}\].
The probability that the ball is red among the six balls is \[\dfrac{1}{6}\] and not being the red ball is \[\dfrac{5}{6}\].
Suppose M is an event where a red ball is drawn and both said the truth.
So,
\[P(M)=P(A)\cdot P(B)\cdot \dfrac{1}{6}=\dfrac{2}{3}\cdot \dfrac{4}{5}\cdot \dfrac{1}{6}=\dfrac{4}{45}\].
Now let N be an event where a non-red ball is drawn and both lie.
Here is a trick in this case. As there are 6 balls so picking one ball except red will leave 5 balls behind. It means that for one ball \[\dfrac{1}{5}\] is associated with A and for B for their lie.
So,\[P(N)=\dfrac{5}{6}\cdot \left( P(A')\cdot \dfrac{1}{5} \right)\cdot \left( P(B')\cdot \dfrac{1}{5} \right)=\dfrac{5}{6}\cdot \dfrac{1}{3}\cdot \dfrac{1}{5}\cdot \dfrac{1}{5}\cdot \dfrac{1}{5}=\dfrac{1}{450}\].
So now by law of probability the truth of the statement is P(E) which is
\[P(E)=\dfrac{P(M)}{P(M)+P(N)}=\dfrac{\dfrac{4}{45}}{\dfrac{4}{45}+\dfrac{1}{450}}=\dfrac{\dfrac{4}{45}}{\dfrac{41}{450}}=\dfrac{40}{41}\].
Hence the required probability is \[\dfrac{40}{41}\].
Note: From Baye’s theorem, this problem reduces to a simple one. Conditional probability will follow here. A and B are independent here so we can determine the truthiness of the statement otherwise it would not be possible to tell the value.
Complete step-by-step answer:
Let the probability of the event such that A speaks the truth is P(A) and the probability of the event such that B speaks the truth is P(B).
Similarly, the probability of the event such that A does not speak the truth is P(A’) and the probability of the event such that B does not speak the truth is P(B’).
So here we are given,
\[P(A)=\dfrac{2}{3}\],\[P(B)=\dfrac{4}{5}\],\[P(A')=1-P(A)=\dfrac{1}{3}\],\[P(B')=1-P(B)=\dfrac{1}{5}\].
The probability that the ball is red among the six balls is \[\dfrac{1}{6}\] and not being the red ball is \[\dfrac{5}{6}\].
Suppose M is an event where a red ball is drawn and both said the truth.
So,
\[P(M)=P(A)\cdot P(B)\cdot \dfrac{1}{6}=\dfrac{2}{3}\cdot \dfrac{4}{5}\cdot \dfrac{1}{6}=\dfrac{4}{45}\].
Now let N be an event where a non-red ball is drawn and both lie.
Here is a trick in this case. As there are 6 balls so picking one ball except red will leave 5 balls behind. It means that for one ball \[\dfrac{1}{5}\] is associated with A and for B for their lie.
So,\[P(N)=\dfrac{5}{6}\cdot \left( P(A')\cdot \dfrac{1}{5} \right)\cdot \left( P(B')\cdot \dfrac{1}{5} \right)=\dfrac{5}{6}\cdot \dfrac{1}{3}\cdot \dfrac{1}{5}\cdot \dfrac{1}{5}\cdot \dfrac{1}{5}=\dfrac{1}{450}\].
So now by law of probability the truth of the statement is P(E) which is
\[P(E)=\dfrac{P(M)}{P(M)+P(N)}=\dfrac{\dfrac{4}{45}}{\dfrac{4}{45}+\dfrac{1}{450}}=\dfrac{\dfrac{4}{45}}{\dfrac{41}{450}}=\dfrac{40}{41}\].
Hence the required probability is \[\dfrac{40}{41}\].
Note: From Baye’s theorem, this problem reduces to a simple one. Conditional probability will follow here. A and B are independent here so we can determine the truthiness of the statement otherwise it would not be possible to tell the value.
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