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(ii)
A ∩ B = Φ
(ii) Let A and B respectively denote the events that both children are females and the elder child is a female.
Using binomial distribution, the probability that more than 6 people are right-handed is given by,
Therefore, the probability that at most 6 people are right-handed
= 1 − P (more than 6 are right-handed)
Question
6. In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6 What is the probability that he will knock down fewer than 2 hurdles?
Solution :
Let
p
and
q
respectively be the probabilities that the player will clear and knock down the hurdle.
Let X be the random variable that represents the number of times the player will knock down the hurdle.
Therefore, by binomial distribution, we obtain
Question
7. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
Solution :
The probability of getting a six in a throw of die is 1/6 and not getting a six is 5/6.
Let p = 1/6 and q = 5/6
The probability that the 2 sixes come in the first five throws of the die is
∴ Probability that third six comes in the sixth throw =
Question
8. If a leap year is selected at random, what is the change that it will contain 52 Tuesday?
Solution :
A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday then the year will have 53 Tuesdays.
Number of total days in a week = 7
Number of favourable days = 2
Therefore, P (the year will have 53 Tuesday) = 2/7
Question
9. An experiment succeeds twice as often as it fails. Find the probability that in the next six trails, there will be at least 4 successes.
Solution :
The probability of success is twice the probability of failure.
Let the probability of failure be
x
.
∴ Probability of success = 2
x
Let X be the random variable that represents the number of successes in six trials.
By binomial distribution, we obtain
Question
10. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
Solution :
Let the man toss the coin
n
times. The
n
tosses are
n
Bernoulli trials.
Probability (
p
) of getting a head at the toss of a coin is 1/2.
∴
p
= 1/2 ⇒
q
= 1/2
It is given that,
P (getting at least one head) > 90/100
P (
x
≥ 1) > 0.9
⇒ 1 − P (
x
= 0) > 0.9
The minimum value of
n
that satisfies the given inequality is 4.
Thus, the man should toss the coin 4 or more than 4 times.
Question
11. In a game a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amounts he wins/looses.
Solution :
When a die is thrown, then probability of getting a six = 16
then, probability of not getting a six = 1 - 16 = 56
If the man gets a six in the first throw, then
probability of getting a six = 16
If he does not get a six in first throw, but gets a six in second throw, then
probability of getting a six in the second throw = 56×16 = 536
If he does not get a six in the first two throws, but gets in the third throw, then
probability of getting a six in the third throw = 56×56×16 = 25216
probability that he does not get a six in any of the three throws = 56×56×56 = 125216
In the first throw he gets a six, then he will receive Re 1.
If he gets a six in the second throw, then he will receive Re (1 - 1) = 0
If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1),
that means he will lose Re 1 in this case.
Expected value = 16×1 + 56×16 × 0 + 56×56×16×-1 = 11216
So, he will loose Rs 11216.
Question
12. Suppose we have four boxes A, B, C and D containing coloured marbles as given below:
Probability that the red marble is from box B is P (E
B
|R).
Probability that the red marble is from box C is P (E
C
|R).
Question
13. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.
Solution :
A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.
Let these events be denoted by E
1
and E
2
i.e.,
E
1
= Treatment of yoga and meditation
E
2
= Treatment of prescription of certain drugs
P (E
1
) = P (E
2
) = 1/2
Let A denotes that a person has heart attack, then P (A) = 40% = 0.40
Yoga and meditation reduces heart attack by 30.
Inspite of getting yoga and meditation treatment heart risk is 70% of 0.40
P (A|E
1
) = 0.40 x 0.70 = 0.28
Also, Drug prescription reduces the heart attack rick by 25%
Even after adopting the drug prescription hear rick is 75% of 0.40
P (A|E
2
) = 0.40 x 0.75 = 0.30
Question
14. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).
Solution :
There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.
Number of determinants that can be formed = (2)
4
= 16
The value of determinants is positive in the following cases:
Therefore, the probability that the determinant is positive = 3/16
Question
15.
An electronic assembly consists of two sub-systems say A and B. From previous testing procedures, the following probabilities are assumed to be known:
P (A fails) = 0.2
P (B fails alone) = 0.15
P (A and B fail) = 0.15
Evaluate the following probabilities.
Solution :
Let the event in which A fails and B fails be denoted by E
A
and E
B
.
P (E
A
) = 0.2
P (E
A
∩ E
B
) = 0.15
P (B fails alone) = P (E
B
) − P (E
A
∩ E
B
)
⇒ 0.15 = P (E
B
) − 0.15
⇒ P (E
B
) = 0.3
(ii) P (A fails alone) = P (E
A
) − P (E
A
∩ E
B
)
= 0.2 − 0.15
= 0.05
Question
16. Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.
Solution :
Let E
1
= Ball transferred from Bag I to Bag II is red
E
2
= Ball transferred from Bag I to Bag Ii is black
A = Ball drawn from Bag II is red in colour
P (E
1
) =3/7 and P (E
2
) = 4/7
Let A be the event that the ball drawn is red.
When a red ball is transferred from bag I to II,
P (A | E
1
) = 5/10 = 1/2
When a black ball is transferred from bag I to II,
P (A | E
2
) = 4/10 = 2/5
Therefore, option (A) is correct.
Question
18.
If P (A|B) > P (A), then which of the following is correct:
Therefore, option (C) is correct.
Question
19. If A and B are any two events such that P (A) + P (B) – P (A and B) = P (A), then
(A) P (B|A) = 1
(B) P (A|B) = 1
(C) P (B|A) = 0
(D)P (A|B) = 0
Solution :
Therefore, option (B) is correct.
