Tag: some more terms in probability

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Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.

  1. $\dfrac57$
  2. $\dfrac{10}{21}$
  3. $\dfrac27$
  4. $\dfrac{11}{21}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total balls = 2 red + 3 green + 2 blue = 7. Total ways to draw 2 balls = 7C2 = 21. Favorable outcomes (no blue) = drawing from 5 non-blue balls = 5C2 = 10. Probability = 10/21.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If the chance that a vessel arrives safely at a port is $\dfrac 9{10}$ then what is the chance that out of $5$ vessels expected at least $4$ will arrive safely?

  1. $\dfrac {14 \times 9^4}{10^5}$
  2. $\dfrac {15 \times 9^5}{10^4}$
  3. $\dfrac {14 \times 9^3}{10^4}$
  4. $\dfrac {14 \times 9^6}{10^5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a binomial distribution problem where n=5, p=0.9, q=0.1. P(X>=4) = P(X=4) + P(X=5). P(X=4) = 5C4 * (0.9)^4 * (0.1)^1 = 5 * 0.9^4 * 0.1 = 0.5 * 9^4 / 10^4 = 5 * 9^4 / 10^5. P(X=5) = 5C5 * (0.9)^5 = 9^5 / 10^5. Sum = (5 * 9^4 + 9 * 9^4) / 10^5 = 14 * 9^4 / 10^5.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

A box contains nine bulbs out of which $4$ are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.

  1. $\dfrac{62}{63}$
  2. $\dfrac{125}{126}$
  3. $\dfrac{1}{63}$
  4. $\dfrac{1}{126}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The total number of ways to choose 4 bulbs from 9 is 9C4 = 126. The number of ways to choose 4 defective bulbs from the 4 available is 4C4 = 1. Thus, the probability is 1/126.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

A pot has $2$ white, $6$ black, $4$ grey and $8$ green balls. If one ball is picked randomly from the pot, what is the probability of it being black or green?

  1. $\dfrac34$
  2. $\dfrac1{10}$
  3. $\dfrac43$
  4. $\dfrac7{10}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total balls = 2 + 6 + 4 + 8 = 20. The number of black or green balls is 6 + 8 = 14. The probability is 14/20, which simplifies to 7/10.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability that A speaks truth is $\dfrac35$ and that of B speaking truth is $\dfrac47$. What is the probability that they agree in stating the same fact?

  1. $\dfrac {18}{35}$
  2. $\dfrac {12}{35}$
  3. $\dfrac {17}{35}$
  4. $\dfrac {19}{35}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

They agree if both tell the truth or both lie. P(Both Truth) = (3/5)(4/7) = 12/35. P(Both Lie) = (2/5)(3/7) = 6/35. Total probability = 12/35 + 6/35 = 18/35.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability of success of three students X,Y and Z in the one examination are $\dfrac15, \dfrac14$ and $\dfrac13$ respectively. Find the probability of success of at least two.

  1. $\dfrac16$
  2. $\dfrac25$
  3. $\dfrac34$
  4. $\dfrac35$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probabilities of success: P(X)=1/5, P(Y)=1/4, P(Z)=1/3. P(Failures): P(X')=4/5, P(Y')=3/4, P(Z')=2/3. At least two success = P(exactly 2) + P(exactly 3). P(exactly 3) = (1/5)(1/4)(1/3) = 1/60. P(exactly 2) = (1/5)(1/4)(2/3) + (1/5)(3/4)(1/3) + (4/5)(1/4)(1/3) = 2/60 + 3/60 + 4/60 = 9/60. Total = 10/60 = 1/6.