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?

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
Advertisements

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

RELATED QUESTIONS

Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find 

  1. P (A and B)
  2. P(A and not B)
  3. P(A or B)
  4. P(neither A nor B)

Two events, A and B, will be independent if ______.


Prove that if E and F are independent events, then the events E and F' are also independent. 


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 at least one of them will be alive 20 years hence.


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


Bag A contains 3 red and 2 white balls and bag B contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, and then a ball is drawn from that bag. Find the probability that both the balls drawn are of same color


Select the correct option from the given alternatives :

The odds against an event are 5:3 and the odds in favour of another independent event are 7:5. The probability that at least one of the two events will occur is


Solve the following:

If P(A ∩ B) = `1/2`, P(B ∩ C) = `1/3`, P(C ∩ A) = `1/6` then find P(A), P(B) and P(C), If A,B,C are independent events.


Solve the following:

A and B throw a die alternatively till one of them gets a 3 and wins the game. Find the respective probabilities of winning. (Assuming A begins the game)


If A and B are independent events such that 0 < P(A) < 1 and 0 < P(B) < 1, then which of the following is not correct?


If A and B are independent events such that P(A) = p, P(B) = 2p and P(Exactly one of A, B) = `5/9`, then p = ______.


For a loaded die, the probabilities of outcomes are given as under:
P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3. The die is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’, respectively. Determine whether or not A and B are independent.


Refer to Question 1 above. If the die were fair, determine whether or not the events A and B are independent.


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


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, B and C are three independent events such that P(A) = P(B) = P(C) = p, then P(At least two of A, B, C occur) = 3p2 – 2p3 


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


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


Let Bi(i = 1, 2, 3) be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α – 2β)p = αβ and (β – 3γ) = 2βy (All the probabilities are assumed to lie in the interval (0, 1)). Then `("P"("B"_1))/("P"("B"_3))` is equal to ______.


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×