## Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Ex 14.3

Question 1.

Probability of event A is 2/11, find the probability of event ‘A-not’.

Solution:

Question 2.

There are 4 male and 6 female members in village paneLIf one member for a committee is selected randomly then what is the probability to select a female.

Solution:

Required probability

Thus, probability, to select a female is \(\frac { 3 }{ 5 } \)

Question 3.

In throwing a dice find the probability of following events.

(i) Appear a prime number

(ii) apear 1 or less than 1

(iii) appear number less than 6.

Solution:

In throwing a dice, obtained sample spapce

S = {1,2, 3, 4, 5, 6}, n(S) = 6

(i) Event E, appear a prime number

then E = {2, 3, 5}

Thus n(E) = 3

Then probability of event E

Thus, probability to get a prime number is = \(\frac { 1 }{ 2 } \)

(ii) Event to get number 1 or less than 1

B = {1}

Then n(B) = 1

Thus, probability to get 1 or less than 1

Thus, probability to get no. 1 or less than 1 = \(\frac { 1 }{ 6 } \)

(iii) Event to appear no. less than 6

D = {1,2, 3,4,5}

then n(D) = 5

Then, probability to appear no. less than 6.

Thus, probability to get number less than 6 = \(\frac { 5 }{ 6 } \)

Question 4.

A coin is tossed 4 times. Find the probability to get tail at least three times in these throw.

Solution:

Sample space obtained by throwing a coin four time

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTP, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

n(S) = 16

Let appearance of tail at least 3 times is event A, then

A = {HHHH, HHHT, HHTH, HTHT, HTHH}

n(A) = 5

Thus, required probability

Thus, probability to get tail at least 3 times is \(\frac { 5 }{ 16 } \)

Question 5.

If a coin and a dice are thrown simultaneously, then what is the probability that tail appear on coin and event number appear on dice ?

Solution:

In throwing coin and dice obtained sample space

S = {H, T} × {1, 2, 3, 4, 5, 6}

= {H1, H2, H3, H4, H5, H6, T1, T2, 73, T4, T5, T6}

n(S) = 12

Let A be the sample space of getting tail on coin and event number on dice, then

A = {H2, H4, H6}

n(A) = 3

Thus, required probability

Thus, required probability is \(\frac { 1 }{ 4 } \).

Question 6.

In a company of 20 people 5 are graduate. If 3 people are selected randomly then what is the probability that one of them be graduate ?

Solution:

Total number of ways in which 3 person are to be select out of 20 = ^{20}C_{3}

Total number of ways to select one graduate = ^{5}C_{2}

and not to select any graduate = ^{15}C_{2}

∴ Total favaurable situations = ^{15}C_{2} × ^{5}C_{1}

Hence, required probability

Question 7.

To solve the problem in opposite of A, odds are 4 : 3. In favour of B odds are 4 : 3. What is the probability that –

(i) Problem wil be solved

(ii) Problem will not be solved

(iii) Problem will be solved only by one

Solution:

Given

(i) Probability of problem solved

(ii) Prob. for problem not solved

(iii) Probability of to be solved by only one

Question 8.

An equipment will work only when its three components A, B and C are in working condition. Probability that A will not work is 0.15 and (M)5 of B and 0.10 to C. What is the probability that equipment will not be in working condition before year ending ?

Solution:

According to question

Probability of damage equipment before year ending

= 1 -P(A).P(B).P(C)

= 1 -0.72675 = 0.27325

Question 9.

From a deck of cards,two cards are randomly drawn one by one. If card once drawn is not replaced then to get two aces in first attempt and two king in second attempt. What is the probability ?

Solution:

Total cards in deck = 52

Probability to get 2 aces in 1st attempt

Probability to get 2 kings in IInd attempt

Question 10.

A and B are two events in which P(A) = \(\frac { 1 }{ 3 } \). P(B) = \(\frac { 1 }{ 4 } \) and P(AB) = \(\frac { 1 }{ 12 } \) then find P(\(\frac { B }{ A } \))

Solution:

Question 11.

Imagine that ratio of male and children are 1:2. Find the probability that in a family out of 5 children are (i) boys (ii) three boys and two girls.

Solution:

Let number of males = x

and Number of boys = x

and Number of girls = x

Probability of choosing boys (P)

(i) Probability of being all boys out of 5 children

(ii) Probability of being 3 boys and 2 girls out of 5 children

Question 12.

A can hit the target 3 times out of 6, 8 can 2 times out of 4 and C can one times out of 5. They let the target simulatenously. What is the probability that at least two persons can hit the target.

Solution:

Given

Probability that at least two person hit the target

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