Q 1 :

Suppose that

Box-I contains 8 red, 3 blue and 5 green balls,
Box-II contains 24 red, 9 blue and 15 green balls,
Box-III contains 1 blue, 12 green and 3 yellow balls,
Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball b. If b is red then a ball is chosen randomly from Box-II, if b is blue then a ball is chosen randomly from Box-III, and if b is green then a ball is chosen randomly from Box-IV. The conditional probability of the event ‘one of the chosen balls is white’ given that the event ‘at least one of the chosen balls is green’ has happened, is equal to                   [2022]

  • 15256    

     

  • 316

     

  • 552

     

  • 18

     

(3)

According to question

[IMAGE 370]

Event E:One of the chosen ball is white

Event F:At least one of the chosen ball is green

 P(EF)=516×63212×1548+316×1216+516×1=552



Q 2 :

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is     [2004]

  • 4/25

     

  • 4/35

     

  • 4/33

     

  • 4/1155

     

(4)

Total numbers which are divisible by both 2 and 3 i.e. 6 are 16

  Probability=C316C3100=41155



Q 3 :

In a study about a pandemic, data of 900 persons was collected. It was found that

190 persons had symptom of fever,
220 persons had symptom of cough,
220 persons had symptom of breathing problem,
330 persons had symptom of fever or cough or both,
350 persons had symptom of cough or breathing problem or both,
340 persons had symptom of fever or breathing problem or both,
30 persons had all three symptoms (fever, cough and breathing problem).

If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is ____________.            [2022]



(0.8)

Let F, C and B be the set of people having symptoms of fever, cough and breathing problem respectively.

Here, we have n(F)=190, n(B)=220 and n(C)=220

Also, n(FC)=330, n(CB)=350, n(FB)=340 and n(FCB)=30

So n(FC)=n(F)+n(C)-n(FC)=80

Similarly, n(FB)=70 and n(CB)=90

By Venn diagram, we can better understand.

Here,

[IMAGE 371]

Number of people having at most one symptom

=70+80+90+480=720

Required probability=720900=0.8