Question

.A box contains 1 − 50 cards. A card is drawn at random. Find the prob. of getting the

card is

a) a prime number

b) multiple of 2 or 5

c) multiple of 2 and 5

d) multiple of 55​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

A box contains 1 − 50 cards. A card is drawn at random

TO DETERMINE

The probability of getting the card is

card isa) a prime number

card isa) a prime numberb) multiple of 2 or 5

card isa) a prime numberb) multiple of 2 or 5c) multiple of 2 and 5

card isa) a prime numberb) multiple of 2 or 5c) multiple of 2 and 5d) multiple of 55

CALCULATION

A box contains 1 − 50 cards

SO total number of cards = 50

A card is drawn at random.

So the total number of possible outcomes = 50

a)

Let A be the event that the card is a PRIME NUMBER

Now the prime numbers from 1 to 50 are

2, 3, 5, 7 , 11 , 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

So total number of primes between 1 and 50 are 15

So the total number of possible outcomes for the event A is 15

So the required probability

$\displaystyle{ \sf{ P(A) = \frac{15}{50} = \frac{3}{10} }}$

b)

Let B be the event that the the number is

MULTIPLE OF 2

So the total number of possible outcomes for the event B is

$= \displaystyle \: \sf{ \bigg[ \frac{50}{2} \bigg] = \displaystyle \: \sf{ \bigg[ 25 \bigg] = 25 \: } \: }$

$\displaystyle \: \sf{ \: P(B) = \frac{25}{50} = \frac{1}{2} \: }$

Let D be the event that the the number is

MULTIPLE OF 5

So the total number of possible outcomes for the event D is

$= \displaystyle \: \sf{ \bigg[ \frac{50}{5} \bigg] = \displaystyle \: \sf{ \bigg[ 10 \bigg] = 10\: } \: }$

$\displaystyle \: \sf{ \: P(D) = \frac{10}{50} = \frac{1}{5} \: }$

Now a number which is multiple of 2 and 5 both is also a multiple of 10

Now

$\sf{B \cap \: D}$

= the event that the the number is multiple of both 2 and 5

So the total number of possible outcomes for the event is

$= \displaystyle \: \sf{ \bigg[ \frac{50}{10} \bigg] = \displaystyle \: \sf{ \bigg[ 5 \bigg] = 5 \: }\: }$

$\displaystyle \: \sf{ \: P(B \cap \: D) = \frac{5}{50} = \frac{1}{10} \: }$

So the required probability

$\displaystyle \: \sf{ \: P(B \: \cup \: D) \: }$

$= \displaystyle \: \sf{ \: P(B) + P(D) -P(B \cap \: D) \: }$

$\displaystyle \: \sf{ \: = \frac{1}{2} + \frac{1}{5} - \frac{1}{10} }$

$\displaystyle \: \sf{ \: = \frac{5 + 2 - 1}{10} }$

$\displaystyle \: \sf{ \: = \frac{6}{10} }$

$\displaystyle \: \sf{ \: = \frac{3}{5} }$

c) Solved in problem ( b)

d)

Let E be the event that the the number is A MULTIPLE OF 55

There is no multiple of 55 between 1 and 50

So the total number of possible outcomes for the event E is 0

Since this is an IMPOSSIBLE EVENT

So the required probability is

$\sf{P(E) = 0}$

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