can any one pls solve all these questions step by step.
Answers
Hey ARMY
37)
prime numbers
38)
a red card
39)
The bag contains only lemon flavored candies, and nothing else.
There are no orange flavored candies in the bag. Hence there is no possibility of taking out an orange candy.
Therefore, the probability of taking out an orange flavored candy =0
Solution(ii):
The bag contains only lemon flavored candies, and nothing else.
Therefore (No.of favorable outcomes)=(Total no.of possible outcomes)
We know that, Probability of an event E, P(E) =
(Total no.of possible outcomes)
(No.of favorable outcomes)
P(E)=1
Therefore, the probability of taking out a lemon flavored candy =1
40)
Let E be the event of having the same birthday.P(E) = 0.992⇒ P(E) + P(not E) = 1⇒ P(not E) = 1 – P(E)⇒ 1 - 0.992 = 0.008The probability that the 2 students have the same birthday is 0.008
41)No. of red balls = 3
No. of black balls = 5
Total no. of balls = 5+3 = 8
(i) Probability of drawing red balls = No. of red balls/Total no. of balls = 3/8
(ii) Probability of drawing black balls = No. of black balls/Total no. of balls = 5/8
42)
total marbles =red +white + green
=5+8+4 =17
(a)red =5/17
(b)white =8/17
(c)not green = red =white
= 5+8 =13
= 13/17
43)
i) P(Defective bulb)=
Total Bulbs
Defective Bulbs =
20
4
= 15
ii) Since one bulb is already drawn, the number of bulbs remaining is 19.
Number of non defective bulbs is 15 as one bulb is picked.
P(Non-defective)= 19/15
44)
i) Let A be the probability getting a 2 digit number
therefore
P(A) = 81/90 = 9/10
ii) Let B be the probability of getting a square number
therefore,
P(B) = 9/90 = 1/10
45)
Given that total number of pens n(S) = 144.
Given that total number of defective pens = 20.
That means total number of non-defective pens = 144 - 20
= 124.
(1) She will buy:
Let A be event that she will buy a non-defective pen
n(A) = .124.
Therefore the required probability P(A) = n(A)/n(S)
= 124/144
= 31/36.
(2) She will not buy it.
Let B be the event that of getting a defective pen.
n(B) = 20. = 5/36.
46)
No.of good pens=156
No.of defective pens=12
Total pens=168
P(E)=no.of favourable outcomes/total no.of outcomes
P(good pen)=156/168
P(good pen)=13/14
Hence probability of good pen is 13/14.
A defective pen is kept aside.
Now,
no.of good pens=156
no.of defective pens=12-1=11
Total no.of pens==167
P(pen is not defective)=no.of good pens/total pens
P(pen is not defective)=156/167
Hence probability of pen not defective is 156/167.
Value shown by the shopkeeper is - whether the pen is good or not.
47)
I)=probability of getting queen=1/5
ll) a=1/4 because only 4 cards are left b=0
48)
total no of possible outcome =4
total no of probable outcome =3
probability =3/4
49)
_________
When we tossed a die , than total number of events ( outcomes ) = 6 , As : { 1 , 2 , 3 , 4 , 5 , 6 }
But when we tossed two dice simultaneously , than total number of events ( outcomes ) = 6 × 6 = 36
As we can show all the possible outcomes , As : { 1 , 1 } , { 1 , 2 } , { 1 , 3 }, { 1 , 4 } ,{ 1 , 5 } , { 1 , 6 } , { 2 , 1 } , { 2 , 2 } , { 2 , 3 }, { 2 , 4 } ,{ 2 , 5 } , { 2 , 6 }, { 3 , 1 } ,{ 3 , 2 }, { 3 , 3 } , { 3 , 4 } ,{ 3 , 5 } , { 3 , 6 } , { 4 , 1 } ,{ 4 , 2 }, { 4 , 3 } , { 4 , 4 } ,{ 4 , 5 } , { 4 , 6 } , { 5 , 1 } ,{ 5 , 2 }, { 5 , 3 } , { 5 , 4 } ,{ 5 , 5 } , { 5 , 6 } , { 6 , 1 } ,{ 6 , 2 }, { 6 , 3 } , { 6 , 4 } ,{ 6 , 5 } , { 6 , 6 }
So,
n ( S ) = 36
We know that
==> probability P ( E ) = Total number of desired events n ( E ) /Total number of events n ( S )
a) The probability that the sum of two numbers appearing on the top of the dice is 8
So,
Number of events = { 2 , 6 }, { 3 , 5 }, { 4 , 4 }, { 5 , 3 } ,{ 6 , 2 }
So,
n ( E ) = 5
The probability that the sum of two numbers appearing on the top of the dice is 8 = 5/36 Ans
____________________
b ) The probability that the sum of two numbers appearing on the top of the dice is 13
So,
Number of events = 0
Then,
n ( E ) = 0
The probability that the sum of two numbers appearing on the top of the dice is 13= 0/36= 0 Ans
____________________
c) The probability that the sum of two numbers appearing on the top of the dice is less than or equal to 12
So,
Number of events = { 1 , 1 } , { 1 , 2 } , { 1 , 3 }, { 1 , 4 } ,{ 1 , 5 } , { 1 , 6 } , { 2 , 1 } , { 2 , 2 } , { 2 , 3 }, { 2 , 4 } ,{ 2 , 5 } , { 2 , 6 }, { 3 , 1 } ,{ 3 , 2 }, { 3 , 3 } , { 3 , 4 } ,{ 3 , 5 } , { 3 , 6 } , { 4 , 1 } ,{ 4 , 2 }, { 4 , 3 } , { 4 , 4 } ,{ 4 , 5 } , { 4 , 6 } , { 5 , 1 } ,{ 5 , 2 }, { 5 , 3 } , { 5 , 4 } ,{ 5 , 5 } , { 5 , 6 } , { 6 , 1 } ,{ 6 , 2 }, { 6 , 3 } , { 6 , 4 } ,{ 6 , 5 } , { 6 , 6 }
So,
n ( E ) = 36
The probability that the sum of two numbers appearing on the top of the dice is less than or equal to 12 = 36/36 = 1 Ans
____________________
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