Skip to content
Question of 165

Q.State Bayes' theorem on probability. Use this theorem to solve the following : An insurance company insured 2000 scooty drivers, 4000 taxi drivers and 6000 bus drivers in a particular year. The probability of their accidents are 0.01, 0.03 and 0.15 respectively. One of the insured drivers meets with an accident. What is the probability that the person drives a scooty? OR State Bayes' theorem on probability. Use this theorem to solve the following : First bag contains 3 red and 4 black balls, and second bag contains 5 red and 6 black balls. A ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from the second bag.

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2021Subjective· 6mImportance★★★★★
0% · 0/165 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 →

State Bayes' theorem, define the partition events (driver/bag type) and the given conditional probabilities, then compute the posterior probability using the theorem.

Bayes' Theorem: If E1,E2,…,EnE_1,E_2,\dots,E_n form a partition of the sample space (mutually exclusive, exhaustive, each with P(Ei)>0P(E_i)>0), and AA is any event with P(A)>0P(A)>0, then for each ii:

P(Ei∣A)=P(Ei) P(A∣Ei)∑j=1nP(Ej) P(A∣Ej)P(E_i\mid A)=\dfrac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^n P(E_j)\,P(A\mid E_j)}

Application: Let E1,E2,E3E_1,E_2,E_3 be the events that a randomly chosen insured driver is a scooty, taxi, or bus driver respectively. Total drivers =2000+4000+6000=12000=2000+4000+6000=12000.

P(E1)=200012000=16,P(E2)=400012000=13,P(E3)=600012000=12P(E_1)=\frac{2000}{12000}=\frac16,\quad P(E_2)=\frac{4000}{12000}=\frac13,\quad P(E_3)=\frac{6000}{12000}=\frac12

Let AA = event that the insured driver meets with an accident. Given:

P(A∣E1)=0.01=1100,P(A∣E2)=0.03=3100,P(A∣E3)=0.15=15100P(A\mid E_1)=0.01=\frac1{100},\quad P(A\mid E_2)=0.03=\frac3{100},\quad P(A\mid E_3)=0.15=\frac{15}{100}

By Bayes' theorem:

P(E1∣A)=P(E1)P(A∣E1)P(E1)P(A∣E1)+P(E2)P(A∣E2)+P(E3)P(A∣E3)P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}

Numerator: 16×1100=1600\dfrac16\times\dfrac1{100}=\dfrac1{600}

Denominator terms: 16×1100=1600\dfrac16\times\dfrac1{100}=\dfrac1{600},  13×3100=1100=6600\ \dfrac13\times\dfrac3{100}=\dfrac1{100}=\dfrac6{600},  12×15100=340=45600\ \dfrac12\times\dfrac{15}{100}=\dfrac{3}{40}=\dfrac{45}{600}

Sum =1+6+45600=52600=\dfrac{1+6+45}{600}=\dfrac{52}{600}

P(E1∣A)=1/60052/600=152P(E_1\mid A)=\frac{1/600}{52/600}=\frac1{52}

…

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.