PROBABILITY WORD PROBLEMS FOR GRADE 10

Problem 1 :

If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.

Solution :

Sample space  =  {(1, 1) (1, 2) (1, 3) .........(6, 6)}

Let "A" be the even of getting the product of face value 6.

A = {(1, 6)(2, 3)(3, 2)(6, 1)}

n(A)  =  4

P(A)  =  n(A)/n(S)  =  4/36

Let "B" be the event of getting the difference of face value as 5.

B = {(1, 6)(6, 1)}

n(B)  =  2

P(B)  =  n(B)/n(S)  =  2/36

A n B  =  {(1, 6)(6, 1)}

P(A n B)  =  2/36

Let "AUB" be the event of getting  the product of face value 6 or the difference of face values 5.

P(A U B)  =  P(A) + P(B) - P(A n B)

  =   (4/36) + (2/36) - (2/36)

  =  4/36

  P(A U B)  =  1/9

Problem 2 :

In a two children family, find the probability that there is at least one girl in a family.

Solution :

Sample space  =  {BB, GG, BG, GB}

n(S)  =  4

Let "A" be the event of getting atleast one girl in the family.

A  =  {GG, BG, GB}

n(A)  =  3

P(A)  =  n(A)/n(S)  

P(A)  =  3/4

Problem 3 :

A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.

Solution :

Let "x" be the number of black balls.

Number of balls  =  5(white) + x(black)

n(S)  =  5 + x

Let "A" and "B" be the events of getting white and black balls respectively.

P(A)  =  5/(5 + x)

P(B)  =  x/(5 + x)

P(B)  =  2P(A)

x/(5+x)  =  2(5/(5+x))

x  =  10

So, the number of black balls is 10.

Problem 4 :

The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?

Solution :

Let "A" and "B" be the probabilities of getting pass marks in English and Tamil.

P(A n B)  =  0.5

P(A bar n B bar)  =  0.1 = P((AUB) whole bar)

P(AUB)  =  1 - P((AUB) whole bar)

  =  1 - 0.1

P(AUB)  =  0.9

P(B)  =  0.75

P(A)  = ?

P(A n B)  =  P(A) + P(B) - P(AUB)

  0.5  =  P(A) + 0.75 - 0.9

  0.5 + 0.15  =  P(A) 

P(A)  =  0.65

  =  65/100

P(A)  =  13/20

Problem 5 :

The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting (i) a diamond (ii) a queen (iii) a spade (iv) a heart card bearing the number 5.

Solution :

Here, King, queen and jack of club are removed from the Deck of 52 playing cards.

So remaining cards in deck= 52 - 3 = 49

Total number of outcomes= 49

(i) we know that there are 13 cards of heart.

Let A = event of getting a heart

P(A) = 13/49

(ii)

We know that there are 4 Kings in a deck. After remove removing a king of club we left with  3 kings. 

Let B = event of getting a king

Number of favourable outcomes to  B  =  3

Required probability P(B)  =  3/49

(iii)

We know that there are 13 cards of club. After removing king ,queen and jack of club only 10 club cards are left in the deck.

Let C = event of getting a club

Number of favourable outcomes to  C = 10

Required probability P(C) = 10/49

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