Case study (4 Marks)

Question 14 Marks

Read the following text and answer the following questions on the basis of the same:
Today in mathematics class Mr. Lal teaches the topic of Binomial distribution. Four friends Rohan, Rohit, Roshan and Raman have few doubts in this topic, so they decided for group study at Rohan's place. They are trying to solve a question in which a pair of dice is thrown 7 times. If getting a total 7 is considered a success, then answer the following questions:

Q.1. If X denote the number of success in 7 throws of a pair of dice, then X Binomial variate with parameters:
(A) $n=7, p=\frac{1}{6}\quad$ (B) $n=2, p=\frac{1}{6}\quad$ (C) $n=7, p=\frac{5}{6}\quad$ (D) $n=2, p=\frac{5}{6}$
Q.2. What is the probability of no success?
(A) $\left(\frac{6}{5}\right)^7\quad$ (B) $\left(\frac{5}{6}\right)^7\quad$ (C) $\left(\frac{1}{5}\right)^7\quad$ (D) $\left(\frac{5}{6}\right)\quad$
Q.3. What is the probability of 6 success?
(A) $\left(\frac{1}{6}\right)^7\quad$ (B) $35\left(\frac{1}{6}\right)\quad$ (C) $35\left(\frac{1}{6}\right)^7\quad$ (D) $35\left(\frac{5}{6}\right)^7\quad$
Q.4. What is the probability of at least 6 success?
(A) $\left(\frac{1}{6}\right)^5\quad$ (B) $\left(\frac{1}{6}\right)^7\quad$ (C) $\left(\frac{1}{6}\right)^6\quad$ (D) $\left(\frac{5}{6}\right)^5$

Answer

(1) (A) $n=7, p=\frac{1}{6}$
Explanation: Let p denote the probability of getting a total of 7 in a single throw of a pair of dice. Then $p=\frac{6}{36}=\frac{1}{6}$
$[\because$ The sum can be 7 in any one of the ways : $(1,6)$, $(6,1),(2,5),(5,2),(3,4)$ and $(4,3)]$
So, the parameters are:
$n=7$ and $p=\frac{1}{6}$
(2) (B) $\left(\frac{5}{6}\right)^7$
Explanation: The X is a binomial variate with parameters $n = 7$ and $p = \frac{1}{6}$
$P(X=r)={ }^7 C_r\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{7-r}$
$r=0,1,2, \ldots,\quad\quad\quad\ldots\text{(i)}$
$\left[\because q=1-p=1-\frac{1}{6}=\frac{5}{6}\right]$
Probability of no success = P(X = 0)
$={ }^7 C_0\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{7-0}$
$=\left(\frac{5}{6}\right)^7$
(3) (C) $35\left(\frac{1}{6}\right)^7$
Explanation: From eq. (i),
Probability of 6 success $= P(X = 6)$
$\begin{array}{l}={ }^7 C_6\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^{7-6} \\ =35\left(\frac{1}{6}\right)^7\end{array}$
(4) (A) $\left(\frac{1}{6}\right)^5$
Explanation: Probability of at least 6 success
$\begin{array}{l}=P(X \geq 6) \\ =P(X=6)+P(X=7) \\ ={ }^7 C_6\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^{7-6}+{ }^7 C_7\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^{7-7} \\ ={ }^7 C_6\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)+{ }^7 C_7\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^0 \\ =7\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)+\left(\frac{1}{6}\right)^7 \\ =\left(\frac{1}{6}\right)^6\left(\frac{35}{6}+\frac{1}{6}\right)=\left(\frac{1}{6}\right)^5\end{array}$

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Question 24 Marks

Read the following text and answer the following questions on the basis of the same:
We take 8 identical slips of paper; write the number 0 on one of them, the number 1 on three of the slips, the number 2 on three of the slips and the number 3 on one of the slip. These slips are folded, put in a box and thoroughly mixed. One slip is drawn at random from the box. If X is the random variable denoting the number written on the drawn slip, answer the following questions:

Q.1. How many total values X can take?
(A) 1 $\quad$ (B) 2 $\quad$ (C) 3 $\quad$ (D) 4
Q.2. Find the probability when X = 0.
(A) $\frac{1}{8}$ $\quad$ (B) $\frac{3}{8}$ $\quad$ (C) $\frac{1}{2}$ $\quad$ (D) $\frac{1}{4}$
Q.3. Find the probability when X = 1.
(A) $\frac{1}{8}$ $\quad$ (B) $\frac{3}{8}$ $\quad$ (C) $\frac{1}{2}$ $\quad$ (D) $\frac{1}{4}$
Q.4. The probability distribution is given as:

Answer

(1) (D) 4
Explanation: Clearly X can take values 0, 1, 2, 3. So, in total X can take total 4 values.
(2) (A) $=\frac{1}{8}$
Explanation: P(X = 0) = (Probability of getting a slip written 0 on it) $=\frac{1}{8}.$
(3) (B) $\frac{3}{8}$
Explanation: P(X = 1) = (Probability of getting a slip written 1 on it) = $\frac{3}{8}.$
(4) (A)

$X$	$0$	$1$	$2$	$3$
$P(X)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Explanation: P(X = 0) = (Probability of getting a slip written 0 on it) $=\frac{1}{8}.$
P(X = 1) = (Probability of getting a slip written 1 on it) $=\frac{3}{8}.$
P(X = 2) = (Probability of getting a slip written 2 on it) $=\frac{1}{8}.$
P(X = 3) = (Probability of getting a slip written 3 on it) $=\frac{1}{8}.$
Thus, probability distribution is given as:

$X$	$0$	$1$	$2$	$3$
$P(X)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

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Question 34 Marks

Read the following text and answer the following questions on the basis of the same:
Mr. Rama Roy a owner of a factory lives in Karnataka. In his factory, he manufactures razor blades. There is a small chance that $\frac{1}{500}$ for any blade to be defective. The blades are in the packets of 10.

Q.1. The probability that packet contain no defective blade is:
(A) $e^{-0.02}$
(B) $0.02$
(C) $0.02~ e^{-0.02}$
(D) $0.002 ~e^{-0.02}$
Q.2. The probability that packet contain one defective blade is:
(A) $e^{-0.02}$
(B) $0.02 e^{-0.02}$
(C) $0.02 e^{-0.002}$
(D) $0.002 e^{-0.02}$
Q.3. Find the probability that packet contain two detective blade, when it is given that $e^{-0.02}=0.9802$
(A) 0.0002 $\quad$ (B) 0.002 $\quad$ (C) 0.2 $\quad$ (D) None of these
Q.4. Find the approximate number of packets containing no defective blade, when there are 10000 packets in a consignment. $\quad\quad (\text{Use }e^{-0.02}=0.9802)$
(A) 9802 $\quad$ (B) 9800 $\quad$ (C) 2 $\quad$ (D) 196

Answer

(1) (A) $e^{-0.02}$
Here given N = 1000 , n = 10
$p=\frac{1}{500}=0.002$
$\Rightarrow \quad m=n p=10 \times 0.002$
$\Rightarrow \quad m=0.02$
$P($ No defective blade $)=P(X=0)$
$=e^{-0.02}(0.002)^0=e^{-0.02}$
(2) (B) $0.02 e^{-0.02}$
Explanation: P(one defective blade) = P(X = 1)
$\begin{array}{l}=\frac{e^{-0.02}(0.02)^1}{1!} \\ =0.02 e^{-0.02}\end{array}$
(3) (A) 0.0002
Explanation: P(two defective blade) = P(X = 2)
$\begin{array}{l}=\frac{e^{-0.02}(0.02)^2}{2!} \\ =\frac{0.9802 \times 0.0004}{2} \\ =0.000196 \\ =0.0002\end{array}$
(4) (A) 9802
Explanation: P (no defective blade) = $e^{-0.02}$
= 0.9802
So, the approximate number of packets containing no defective blade
$= 10000 \times 0.9802 = 9802$

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Applied Maths Case study (4 Marks)

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