3 Marks Each

Question 513 Marks

If $P(B)=\frac{3}{5}$ And $P \left( A ^{\prime} \cap B \right)=\frac{1}{2}$, for two events A and B, find $P\left(\frac{A}{B}\right)$ and $P \left( A ^{\prime} \cup B ^{\prime}\right)$.

Answer

1. $P(A' \cap B) = P(B) - P(A \cap B) \implies \frac{1}{2} = \frac{3}{5} - P(A \cap B) \implies P(A \cap B) = \frac{3}{5} - \frac{1}{2} = \frac{1}{10}$.
2. $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/10}{3/5} = \frac{1}{6}$.
3. $P(A' \cup B') = 1 - P(A \cap B) = 1 - \frac{1}{10} = \frac{9}{10}$.

View full question & answer→

Question 523 Marks

State the characteristics of random experiment.

Answer

1. It is very simple to understand and easy to draw.
2. No mathematical formula or complex calculations are required.
3. It can be used even if the trend is not linear.

View full question & answer→

Question 533 Marks

The details of a sample inquiry of 4979 voters of constituency are as follows :

Details	Males	Females
Supporters of Party A	1319	1118
Supporters of Party B	1217	1325

One voter is randomly selected from this constituency.
(1) If this voter is a male, find the probability that he is a supporter of party A.
(2) If this voter is a supporter of party A, find the probability that he is a male.

Answer

Total Males (M) = 1319 + 1217 = 2536.
Total Supporters of A (S_A) = 1319 + 1118 = 2437.
Intersection (Male and Supporter of A) = 1319.
(1) $P(A|M) = \frac{1319}{2536} \approx 0.52$.
(2) $P(M|A) = \frac{1319}{2437} \approx 0.54$.

View full question & answer→

Question 543 Marks

The probability that the price of potato rises in the vegetable market during festive days is 0.8. The probability that the price of onion rises is 0.7. The probability of rise in price of both potato and onion is 0.6. Find the probability of rise in price of at least one of the two, potato and onion.

Answer

$P(A)=0.8, P(B)=0.7, P(A\cap B)=0.6$.
At least one rises means $P(A\cup B)$.
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
$P(A\cup B) = 0.8 + 0.7 - 0.6 = 0.9$.

View full question & answer→

Question 553 Marks

Find the probability of having 5 Tuesdays in the month of August of any year.

Answer

August has 31 days.
31 days = 4 weeks and 3 extra days.
In 4 weeks, there are 4 Tuesdays.
Total possible outcomes for 3 extra days: (Sun, Mon, Tue), (Mon, Tue, Wed), (Tue, Wed, Thu), (Wed, Thu, Fri), (Thu, Fri, Sat), (Fri, Sat, Sun), (Sat, Sun, Mon). Total = 7.
Favorable outcomes (containing Tue): (Sun, Mon, Tue), (Mon, Tue, Wed), (Tue, Wed, Thu). Fav = 3.
Probability = $\frac{3}{7}$.

View full question & answer→

Question 563 Marks

The probability that the price of potato rises in vegetable market during festival days is 0.8.
The probability that price of onion rises is 0.7. The probability of rise in prices of both potatoes and onions is 0.6. Find the probability of rise in price of at least one of the two potatoes and onions.

Answer

Let $ P(A) = 0.8, P(B) = 0.7, P(A \cap B) = 0.6 $
$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.6 = 0.9 $.

View full question & answer→

Question 573 Marks

Find the probability of having 53 Thursdays in a leap year.

Answer

A leap year has 366 days. $ 366 = 52 $ weeks + 2 days.
These 2 days can be (Mon,Tue), (Tue,Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon).
Total outcomes $ n=7 $. Favorable outcomes for Thursday are (Wed,Thu) and (Thu,Fri), so $ m=2 $.
Probability = $ \frac{2}{7} $.

View full question & answer→

Question 583 Marks

The sample space of a random experiment of selecting a number is $U=\{1,2,3 \ldots ., 20\}$. Write the sets showing the following events :
(i) The selected number is odd number.
(ii) The selected number is divisible by 3.
(iii) The selected number is divisible by 2 or 3.

Answer

(i) $ \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\} $
(ii) $ \{3, 6, 9, 12, 15, 18\} $
(iii) Divisible by 2: $ \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\} $. Divisible by 3: $ \{3, 6, 9, 12, 15, 18\} $.
Union : $ \{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20\} $

View full question & answer→

Question 593 Marks

A problem in Mathematics is given to Dhyan, Kaushal and Namrata to solve the probabilities of them solving the problem correctly are $\frac{2}{3}, \frac{3}{4}$ and $\frac{1}{2}$ respectively.
Find the probability that the problem is solved correctly.

Answer

Student	Success P(A)	Failure P(A')
Dhyan (D)	$2 / 3$	$1-2 / 3=1 / 3$
Kaushal (K)	$3 / 4$	$1-3 / 4=1 / 4$
Namrata (N)	$1 / 2$	$1-1 / 2=1 / 2$

$P($ None solve $)=P\left(D^{\prime}\right) \times P\left(K^{\prime}\right) \times P\left(N^{\prime}\right)$
$=\frac{1}{3} \times \frac{1}{4} \times \frac{1}{2}=\frac{1}{24}$
$P($ Solved $)=1-P($ None solve $)$
$P($ Solved $)=1-\frac{1}{24}$
$P($ Solved $)=\frac{23}{24}$

View full question & answer→

Question 603 Marks

Three events A, B and C in a sample space are mutually exclusive and exhaustive. If $P(C')=0.8$ and $3P(B)=2P(A')$ then find $P(A)$ and $P(B)$.

Answer

$P(C) = 1 - 0.8 = 0.2$.
Since exhaustive, $P(A)+P(B)+P(C) = 1 \implies P(A)+P(B) = 0.8$.
$3P(B) = 2(1 - P(A)) \implies 3P(B) = 2 - 2P(A) \implies 2P(A) + 3P(B) = 2$.
Solving both : $P(A) = 0.4$
$P(B) = 0.4$.

View full question & answer→

Question 613 Marks

Seven sneakers P, D, J, O, S, H, I are invited in a programme to deliver speech in random order. Find the probability that D delivers speech immediately after speaker P .

Answer

Total arrangements $n = 7!$.
Treat (PD) as one block. Arrangements $m = 6!$.
Probability $P = 6!/7! = 1/7$.

View full question & answer→

Statistics 3 Marks Each