English

The odds against student X solving a business statistics problem are 8 : 6 and odds in favour of student Y solving the same problem are 14 : 16 What is the probability that neither solves the problem? - Mathematics and Statistics

Advertisements
Advertisements

Question

The odds against student X solving a business statistics problem are 8: 6 and odds in favour of student Y solving the same problem are 14: 16 What is the probability that neither solves the problem?

Diagram
Sum

Solution

Let A be the event that X solves the problem B be the event that Y solves the problem.
Since the odds against student X solving the problem are 8: 6
∴ Probability of occurrence of event A is given by.

P(A) = `6/(8 + 6) = 6/14` and

P(A') = 1 – P(A) = `1 - 6/14 = 8/14`

Also, the odds in favour of student Y solving the problem are 14: 16
∴ Probability of occurrence of event B is given by

P(B) = `14/(14 + 16) = 14/30` and

P(B') = 1 – P(B) = `1 - 14/30 = 16/30`

Now A and B are independent events.
∴ A' and B' are independent events.

P (neither solves the problem) = P(A' ∩ B')
= P(A') . P(B')

= `8/14 xx 16/30`

= `32/105`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Probability - Miscellaneous Exercise 7 [Page 110]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) [English] 11 Standard Maharashtra State Board
Chapter 7 Probability
Miscellaneous Exercise 7 | Q 14. (b) | Page 110

RELATED QUESTIONS

If `P(A)  = 3/5 and P(B) = 1/5` , find P (A ∩ B) if A and B are independent events.


Events A and B are such that `P(A) = 1/2, P(B) = 7/12 and P("not A or not B") = 1/4` . State whether A and B are independent?


One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third guns respectively. assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.


The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that the couple will be alive 20 years hence.


A, B, and C try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are `3/4, 1/2` and `5/8`. Find the probability that the target

  1. is hit exactly by one of them
  2. is not hit by any one of them
  3. is hit
  4. is exactly hit by two of them

Two hundred patients who had either Eye surgery or Throat surgery were asked whether they were satisfied or unsatisfied regarding the result of their surgery

The follwoing table summarizes their response:

Surgery Satisfied Unsatisfied Total
Throat 70 25 95
Eye 90 15 105
Total 160 40 200

If one person from the 200 patients is selected at random, determine the probability that the person was satisfied given that the person had Throat surgery.


Two hundred patients who had either Eye surgery or Throat surgery were asked whether they were satisfied or unsatisfied regarding the result of their surgery

The follwoing table summarizes their response:

Surgery Satisfied Unsatisfied Total
Throat 70 25 95
Eye 90 15 105
Total 160 40 200

If one person from the 200 patients is selected at random, determine the probability that person was unsatisfied given that the person had eye surgery


Two hundred patients who had either Eye surgery or Throat surgery were asked whether they were satisfied or unsatisfied regarding the result of their surgery.

The following table summarizes their response:

Surgery Satisfied Unsatisfied Total
Throat 70 25 95
Eye 90 15 105
Total 160 40 200

If one person from the 200 patients is selected at random, determine the probability the person had Throat surgery given that the person was unsatisfied.


Solve the following:

Find the probability that a year selected will have 53 Wednesdays


Solve the following:

For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = `2/3`, P(B ∪ C) = `3/4`, P(A ∪ B ∪ C) = `11/12`. Find P(A), P(B) and P(C)


Solve the following:

A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which is able to detect any part that is obviously defective and discard it. What is the quality of the parts that make it throught the inspection machine and get shipped?


The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then prove that P(A′) + P(B′) = 2 – 2p + q.


A and B are two events such that P(A) = `1/2`, P(B) = `1/3` and P(A ∩ B) = `1/4`. Find: `"P"("B"/"A")`


Let E1 and E2 be two independent events such that P(E1) = P1 and P(E2) = P2. Describe in words of the events whose probabilities are: (1 – P1) P2 


Let E1 and E2 be two independent events such that P(E1) = P1 and P(E2) = P2. Describe in words of the events whose probabilities are: 1 – (1 – P1)(1 – P2


Let E1 and E2 be two independent events such that P(E1) = P1 and P(E2) = P2. Describe in words of the events whose probabilities are: P1 + P2 – 2P1P2 


If A and B are two events such that P(B) = `3/5`, P(A|B) = `1/2` and P(A ∪ B) = `4/5`, then P(A) equals ______.


Let A and B be two events such that P(A) = `3/8`, P(B) = `5/8` and P(A ∪ B) = `3/4`. Then P(A|B).P(A′|B) is equal to ______.


Let E1 and E2 be two independent events. Let P(E) denotes the probability of the occurrence of the event E. Further, let E'1 and E'2 denote the complements of E1 and E2, respectively. If P(E'1 ∩ E2) = `2/15` and P(E1 ∩ E'2) = `1/6`, then P(E1) is


Let EC denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1 ∩ E2 ∩ E3) = 0. Then `"P"(("E"_2^"C"  ∩ "E"_3^"C")/"E"_1)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×