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## Class 11 math (India)

### Course: Class 11 math (India)>Unit 15

Lesson 4: Solutions to select NCERT problems

# Select problems from exercise 16.3

Solutions to some problems of NCERT exercise.
In this article we will look at solutions of a few selected problems from exercise 16.3 of NCERT.
Problem 1:
A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs 1, point, 50 for each tail that turns up.
From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Solution:
On each toss the person may either win or lose. The final amount the person gets depends on the number of times he wins or loses.
There are a total of 5 possible cases. Below start text, W, end text and start text, L, end text denote a win and loss respectively.
Case start text, I, end text: 4 wins.
Outcome is start text, W, W, W, W, end text. Here the person wins Rs 4 in total.
Case start text, I, I, end text: 3 wins and 1 loss.
Outcomes are start text, W, W, W, L, end text, start text, W, W, L, W, end text, start text, W, L, W, W, end text and start text, L, W, W, W, end text. Here the person wins Rs 1, point, 50 in total.
Case start text, I, I, I, end text: 2 wins and 2 losses.
Outcomes are start text, W, W, L, L, end text, start text, W, L, W, L, end text etc. Here the person loses Re 1 in total.
Try it out
The bigger question is, how many outcomes are possible for this case?

We can list out all the outcomes but a shortcut would be to use permutation theory.
Basically, here we are arranging 4 objects, 2 identical start text, W, s, end text and 2 identical start text, L, s, end text in a line. Number of ways to do that is start fraction, 4, !, divided by, 2, !, 2, !, end fraction, equals, 6.
Case start text, I, V, end text: 1 win and 3 losses.
Try it out
Using permutation theory, find out the number of outcomes in this case.

There are 4 outcomes: start text, L, L, L, W, end text, start text, L, L, W, L, end text, start text, L, W, L, L, end text and start text, W, L, L, L, end text. Here the person loses Rs 3, point, 50 in total.
Case start text, V, end text: 4 losses.
Outcome is start text, L, L, L, L, end text. Here the person loses Rs 6 in total.
Now how do we find the probability for each case?
start text, P, end text, equals, start fraction, start text, N, o, point, space, o, f, space, f, a, v, o, u, r, a, b, l, e, space, o, u, t, c, o, m, e, s, end text, divided by, start text, T, o, t, a, l, space, n, o, point, space, o, f, space, o, u, t, c, o, m, e, s, end text, end fraction
Total number of outcomes is the sum of number of outcomes in each case equals, 1, plus, 4, plus, 6, plus, 4, plus, 1, equals, 16.
Required probabilities are
CaseResultNo. of favourable outcomesProbability
start text, I, end textWin Rs 41start fraction, 1, divided by, 16, end fraction
start text, I, I, end textWin Rs 1, point, 504start fraction, 1, divided by, 4, end fraction
start text, I, I, I, end textLose Re 16start fraction, 3, divided by, 8, end fraction
start text, I, V, end textLose Rs 3, point, 504start fraction, 1, divided by, 4, end fraction
start text, V, end textLose Rs 61start fraction, 1, divided by, 16, end fraction
Problem 2:
In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game.
[Hint: order of numbers is not important.]
Solution:
What is the number of possible outcomes in this case?
The person can pick any combination of 6 numbers from 20 numbers. Number of possible choices is start superscript, 20, end superscript, start text, C, end text, start subscript, 6, end subscript.
Out of these, only 1 is the winning combination. The person wins if he picks this exact combination.
Probability of winning equals, start fraction, 1, divided by, start superscript, 20, end superscript, start text, C, end text, start subscript, 6, end subscript, end fraction, equals, start fraction, 1, divided by, start fraction, 20, !, divided by, 6, !, 14, !, end fraction, end fraction, equals, start fraction, 1, divided by, 38760, end fraction.
Problem 3:
Events start text, E, end text and start text, F, end text are such that start text, P, left parenthesis, n, o, t, space, E, space, o, r, space, n, o, t, space, F, right parenthesis, space, =, space, 0, point, 25, end text. State whether start text, E, end text and start text, F, end text are mutually exclusive.
Solution:
Mutually exclusive events can never happen simultaneously. In other words, if events start text, E, end text and start text, F, end text are mutually exclusive, start text, P, left parenthesis, E, space, a, n, d, space, F, right parenthesis, end text, equals, start text, P, end text, left parenthesis, start text, E, end text, \cap, start text, F, end text, right parenthesis, equals, 0.
Given start text, P, left parenthesis, n, o, t, space, E, space, o, r, space, n, o, t, space, F, right parenthesis, space, =, space, 0, point, 25, end text.
How can we convert this into an equation about probability of start text, E, space, a, n, d, space, F, end text?
See that
\begin{aligned} \text {not E or not F} &= \text E' \cup \text F' \\\\ \text E' \cup \text F' &= (\text E \cap \text F)' \quad \small \text{using DeMorgan's law} \end{aligned}
Therefore
\begin{aligned} \text P(\text E' \cup \text F') &= \text P(\text E \cap \text F)' \\\\ &= 1 - \text P(\text E \cap \text F) \quad \small \because \text P (\text A') = 1 - \text P (\text A) \end{aligned}
Given start text, P, end text, left parenthesis, start text, E, end text, prime, \cup, start text, F, end text, prime, right parenthesis, equals, 0, point, 75. So
\begin{aligned} 0.75 &= 1 - \text P(\text E \cap \text F) \\\\ \text P(\text E \cap \text F) &= 0.25 \end{aligned}
Because start text, P, end text, left parenthesis, start text, E, end text, \cap, start text, F, end text, right parenthesis, does not equal, 0, start text, E, end text and start text, F, end text are not mutually exclusive events.
Problem 4:
The probability that a student will pass the final examination in both English and Hindi is 0, point, 5 and the probability of passing neither is 0, point, 1. If the probability of passing the English examination is 0, point, 75, what is the probability of passing the Hindi examination?
Solution:
Let us make a Venn diagram of probabilities using given data.
start text, P, left parenthesis, H, i, n, d, i, right parenthesis, end text, equals, 0, point, 5, plus, start text, question mark, end text
Because sum of all probabilities should equal 1, we have
\begin{aligned} 0.25 + 0.5 + \text ? + 0.1 &= 1 \\\\ \text ? &= 0.15 \end{aligned}
Therefore start text, P, left parenthesis, H, i, n, d, i, right parenthesis, end text, equals, 0, point, 5, plus, 0, point, 15, equals, 0, point, 65.

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