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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.

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

In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11 ?

  1. $\displaystyle \frac{7}{12}$
  2. $\displaystyle \frac{5}{36}$
  3. $\displaystyle \frac{1}{6}$
  4. $\displaystyle \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let A denotes the event of getting a total of 10 and B denotes the event of getting a total of 11.
Total outcomes = $6\times 6 $
A= {(5,5),(6,4),(4,6)}
n(A) = $3$
$\therefore $ p(A) = $\dfrac {3}{36}  $
B={(6,5),(5,6)}
n(B) = 2
$\therefore $ p(B)= $\dfrac {2}{36}  $
$\therefore $ p(A or B) = p(A) + p(B) 
                   = $\dfrac {3}{36} + \dfrac {2}{36} $
                   = $\dfrac {5}{36} $
Option B is correct.

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

How many times must a man toss a fair coin, so that the probability of having at least one head is more than $80 \%?$

  1. $3$
  2. $>3$
  3. $<3$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In any fair coin toss, P (getting a head) = P (getting a tail) i.e., p=q=$\dfrac 12$
We need to find n such that the probability of getting at least one head is more than $80\%$
$P(X≥1)=1−P(X<1)>80\%$
$\implies 1−P(X=0)>\dfrac 8{10}\\\implies P(X=0)<1−\dfrac 8{10}\\\implies P(X=0)<\dfrac 2{10} or P(X=0)<\dfrac 15$
For a bionomial distribution, $P(X=0)=^nC _0\left(\dfrac 12\right)^0\left(\dfrac 12\right)^{n−0}=\left(\dfrac 12\right)^n$
$\implies \left(\dfrac 12\right)^n<\dfrac 1{5}\\\implies 2^n>5$
Since $2^1=2,2^2=4, 2^3=8,2^4=16$, the minimum value for n that satisfies the inequality is $n=3$, i.e, the coin should be tossed $3$ or more times.