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

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

Calculate the probability that a spinner, having the numbers one through five evenly spaced, will land on an odd number exactly once if the spinner is used three times.

  1. $\dfrac {12}{125}$
  2. $\dfrac {18}{125}$
  3. $\dfrac {27}{125}$
  4. $\dfrac {36}{125}$
  5. $\dfrac {54}{125}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total possible combinations when spinner used three times is $5 \times 5 \times 5 = 125$.
Out of three times, inexactly one number is odd implies the other two numbers are even.
The possible combinations such that exactly one number is odd is $3 \times 2 \times 2 + 2 \times 3 \times 2 + 2 \times 2 \times 3 = 36$.
The probability is $\dfrac {36}{125}$.

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

A bag contains $10$ balls, each labelled with a different integer from $1$ to $10$, inclusive. If $2$ balls are drawn simultaneously from the bag at random, calculate the probability that the sum of the integers on the balls drawn will be greater than $6$.

  1. $0.41$
  2. $0.43$
  3. $0.60$
  4. $0.76$
  5. $0.87$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Out of $10$ balls , $2$ balls can be selected in ${ ^{ 10 }{ C } } _{ 2 } = 45$
Number of ways of selecting $2$ balls such that sum is less than or equal to $6$ is $6$
Probability that the sum of integers on the balls drawn will be greater than $6$ is $1-\dfrac {6}{45} = \dfrac {39}{45} = 0.87$