Skip to content
Question 71 of 89

Q.(a) There are two identical urns containing respectively 6 black and 4 red balls, 2 black and 2 red balls. An urn is chosen at random and a ball is drawn from it.

(i) Find the probability that the ball is black.
(ii) If the ball is black, what is the probability that it is from the first urn? OR
(b) Rewrite 3x−y+4=0\sqrt{3}x - y + 4 = 0 into normal form.
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2022Subjective· 5mImportance★★★★★
80% · 71/89 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

By total probability, P(black)=1120P(\text{black})=\dfrac{11}{20}; by Bayes' theorem, P(urn 1∣black)=611P(\text{urn 1}\mid\text{black})=\dfrac{6}{11}.

Let U1U_1 = urn 1 chosen (6 black, 4 red, total 10), U2U_2 = urn 2 chosen (2 black, 2 red, total 4). Since the urns are identical and chosen at random, P(U1)=P(U2)=12P(U_1)=P(U_2)=\dfrac12.

P(black∣U1)=610=35P(\text{black}\mid U_1)=\dfrac{6}{10}=\dfrac35; P(black∣U2)=24=12P(\text{black}\mid U_2)=\dfrac{2}{4}=\dfrac12.

(i) By the law of total probability:

P(black)=P(U1)P(black∣U1)+P(U2)P(black∣U2)=12⋅35+12⋅12=310+14=620+520=1120P(\text{black}) = P(U_1)P(\text{black}\mid U_1)+P(U_2)P(\text{black}\mid U_2) = \dfrac12\cdot\dfrac35+\dfrac12\cdot\dfrac12 = \dfrac{3}{10}+\dfrac14 = \dfrac{6}{20}+\dfrac{5}{20}=\dfrac{11}{20}.

(ii) By Bayes' theorem:

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.