JEE Advanced Level Test: Probability- 1 - EmSAT Achieve MCQ

Probability of not getting a head = 1/2
Probability of getting at least one head in n tosses is > 0.8 (given)
⇒ 1 -P (getting no head) > 0.8


Putting values for n, we see that the least value of n satisfying (1) is n=3.

Let S be success &F be failure and let p = probability of getting head = 1/2
q = 1/2
A starts the game then
P (A wins) = P (A wins in first throw) + P (A wins in third throw) + P (A wins in fifth throw) +......... ∞
= P(S) + P(FFS) + P(FFFFS) +......... ∞
= p + q.q.p + q.q.q.q.p +.........


This is an infinite GP and hence its sum is

P (A will die within a year) = x
P (B will die within a year) = y
∴ P (Exactly one of them will be alive at the end of the year)
= P (A is alive & B is dead)+ P (A dies B is alive)

(As these are independent events)

Out of integers from 1 to 11, there are 5 even and 6 odd integers.
Let A – both the numbers chosen are odd
B –sum of the numbers chosen is even. 

Let
X = TATANAGAR
Y = CALCUTTA
P(X) = P(Y) = 1/2
E – Two consecutive letters TA are visible.
Two consecutive letters from X are TA, AT, TA, AN, NA, AG, GA, AR

Two consecutive letters from Y are CA, AL, LC, CU, UT, TT, TA

∴ By Bayes Theorem,

Let X – No. of defective items in a sample of 10 items drawn successively.
Trial of drawing defective items with replacement are Bernoulli trials.
Probability of success = p = 5/100 = 1/20
∴ Probability of failure = 

    (1)
P/not more than 1 defective

Let n (s) = total number of ways of seating (permutations) 6 boys and 6 girls
i.e. 12 people = ¹²P₁₂ = 12!
Let E: six girls sit together.
Then they become one unit
∴ No. of units is 6 boys + 1 unit of girls = 7 units
∴ n (E) = No. of ways in which six girls can sit together is = 7!, 6!
(Here 6! is the number of ways in which the six girls can sit among themselves)

These are seven days in a week and every day can have the chance of being any out of Sunday, Monday, Tuesday, ... Saturday (7 choices)
∴ Reqd. probability = 1/7

n(s) = total number of ways of putting 4 letters in 4 envelopes.
= ⁴P₄ = 4! = 24
No. of ways that all letters are placed in right envelope is 1.
∴ n(E)= No. of ways that all letters are not placed in right envelope.
(i.e. at least one letter is put in wrong envelope) = 24 - 1 = 23.

n(S) = Total number of ways of pulling out socks

Let E be the event that they are of the same colour i.e. either both brown or both white.

There are 3 girls and 5 boys
∴ Total number of boys and girls = 8
n(s) = Total number of ways of choosing 3 persons

Let E be the event that there are more girls than boys.
⇒ (2 girls & 1 boy) or (3 girls & 0 boy)

Remember, if the total number of tickets is (2n + 1) and three (3) tickets are drawn at random; then the number of ways that numbers on the drawn tickets are in AP is n².
Here 2n + 1 = 15
⇒ 2n = 14
⇒ n = 7

Remember that if two people A&B perform an experiment and p is the probability of success in the experiment and q that of failure in the same experiment (and assuming that the game continues indefinitely till one of them meets with success and hence the game); then probability of A & B winning the game are respectively 
If there are 3 people then the probabilities of A, B & C wining the game are respectively

∴ we have the respective probabilities of A, B & C winning as

Hexagon has 6 vertices. Three vertices are chosen.
∴ n(S) = Total number of triangles formed


Let E : Triangle is equilateral
Out of these 20 triangles, only two triangles ACE and BDF are equilateral and each of their side is where ‘a’ is the side of regular hexagon.

n(S) = Total number of ways of selecting 5 tickets out of 50 tickets = ⁵⁰C₅

∴ The first two tickets are from 29 tickets (numbered 1 to 29) and last two x4& x5 are from next 20 (numbered 31 to 50)

Let, probability that C wins = p
Then probability that B wins = 2p
Probability of A wins = 2(2p) = 4p
Now
P(A) + P(B) + P(C) = 1
⇒ 4p + 2p + p = 1
⇒ 7p = 1

P(A speaks truth) = 75% = 75/100 = 3/4
∴  = P(A tells a lie)
P(A speaks truth) = 80% = 80/100 = 4/5
 = P(B tells a lie)
∴ P (A & B contradict each other)
= P (A speaks truth & B tells a lie)
+ P (A tells a lie & B speaks truth)

(∴ A&B are independent events)

Let
A – Getting 15 as the sum in a throw of three dice.
B – Getting 4 on the first die.
Required probability 

(As total number of ways = 1×6×6=36. There are two favourable events (4,6,5) (4,5,6))

Let  A – Getting exactly one head
E₁ – She threw 5 or 6
E₂ – She threw 1, 2, 3 or 4


Now possible outcomes when throwing a coin three times are
{HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}
Exactly one head = {TTH, THT, HTT}

When coin is tossed once, the outcomes are {H, T}

Let  E₁ : Head appears
E₂ : Tail appears
A : A reports that head appears

∴ Rwquired probability = 
By Bayes’ Theorem

Let A – Ball drawn from Bag II is red in colour
E₁ – Ball transferred from Bag I to Bag II is red
E₂ – Ball transferred from Bag I to Bag II is black.



By Bayes theorem

Let D – Lost card is diamond
D – Lost card is not diamond.

When one diamond card is lost, the number of diamond cards are 12 out of 51.
No. of ways of choosing 2 from 12 cards ¹²C₂ .
Similarity, choosing 2 diamond cards out of 51 are ⁵¹C₂.
 probability of getting two cards, when one diamond is lost

When the lost and is not diamond, number of diamond cards are 13 out of 51.
No. of ways of choosing 2 cards when one card is lost which is not diamond =

Let
A – The answer is correct
E1 – Student. Knows the answer
E2 – Student Guesses

 We know that he knows and he answers correctly.
= 1.
 We know that he guesses and the answer is correct.
= 1/4
P(E₁) = He knows the answer = 3/4
P(E₂) = He guesses = 1/4
Substituting all these values in (1) we get :

Let
X – no. of correct answers by guessing in the set of 5 multiple choice questions. 5 multiple choice questions.
Probability of success = p = 1/3
Probability of failure = q = 1 - p = 2/3
P(r) = ⁿC_r p^rq^n-r  r = 0,1,2,,....n
P (guessing more than 4 correct answers)


 

Let H denote the head,T the tail and * any of the head or tail.
Then,P(H) = 1/2, P(T) = 1/2 and P(∗)=1.
For at least four consecutive heads, we should have any of the following patterns:
Probability
i) HHHH*** (1/2)⁴×1=1/16
ii)THHHH ** (1/2)⁵ = 1/32
iii) *THHHH* (1/2)⁵ = 1/32
iv) **THHHH (1/2)⁵ = 1/32
Since, all the above cases are mutually exclusive, 
the probability of getting at least four consecutive heads (on adding) is 1/16 + 3/32 = 5/32

Total number of ways of distribution is 45
∴ n(S) = 4⁵
Total number of ways of distribution so that each child gets at least one game is 
4⁵ − ⁴C₁ 3⁵ + ⁴C₂ 2⁵ − ⁴C₃
= 1024 − 4 × 243 + 6 × 32 − 4 
∴ n(E)=240
Therefore, the required probability is 
n(E)/n(S) = 240/4⁵ 
= 15/64