Maharashtra Board Class 10 Math Part – 1 Solution Chapter 5 Practice Set 5.3 – Probability
Balbharati Maharashtra Board Class 10 Math Part – 1 Solution Chapter 5: Probability. Marathi or English Medium Students of Class 10 get here Probability full Exercise Solution.
| Std | Maharashtra Class 10 |
| Subject | Math Part 1 Solution |
| Chapter | Probability |
| Practice Set | 5.3 |
Practice Set – 5.3
(1) Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
(1) One die is rolled,
Event A: Even number on the upper face.
Event B: Odd number on the upper face.
Event C: Prime number on the upper face.
Solution:
One die is rolled.
S = {1, 2, 3, 4, 5, 6}, n(s) = 6.
For event A:
A = {2, 4, 6} n(A) = 3
For event B:
B = {1, 3, 5} n(B) = 3
For event C:
C = {3, 2, 3, 5} n(C) = 3
(2) Two dice are rolled simultaneously,
Event A: The sum of the digits on upper faces is a multiple of 6.
Event B: The sum of the digits on the upper faces is minimum 10.
Event C: The same digit on both the upper faces.
Solution:
Two dies are rolled
S = {(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)}
n(s) = 36
For event A:
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
n(A) = 6
For event B:
B = {(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)}
n (B) = 33
For event C:
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n (C) = 6
(3) Three coins are tossed simultaneously.
Condition for event A: To get at least two heads.
Condition for event B: To get no head.
Condition for event C: To get head on the second coin.
Solution:
3 coin tossed simultaneously.
S {HHH, HHT, HTT, HTH, TTT, TTH, THH, THT}
n (s) = 8
For event A:
A = {HHH, HHT, HTH, THH}
n (A) = 4
For event B:
B = {TTT}
n (n) = 1
For even C:
C = {HHH, HHT, THH, THT}
n(c) = 4
(4) Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A: The number formed is even
Condition for event B: The number formed is divisible by 3.
Condition for event C: The number formed is greater than 50.
Solution:
Two digit number formed using 0, 1, 2, 3, 4, 5. Without-repetition.
S = {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54}
n (s) = 25
For event – A:
A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}
n (A) = 13
For even B:
B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
n (B) = 9
For event C:
C = {51, 52, 53, 54}
n (c) = 4
(5) From three men and two women, environment committee of two persons is to be formed.
Condition for event A: There must be at least one woman member.
Condition for event B: One man, one woman committee to be formed.
Condition for event C: There should not be a woman member.
Solution:
From 3 men two women environment committee of two person is formed.
Let, 3 men be M1, M2, M3
2 women be w1, w2
∴ S = {M1, M2), (M1, M3), (M1, W1), (M1, W2), (M2, M3), (M2, W1), (M3, W1), (M2, W2), (W1, W2)}
n(s) = 10
For event A:
A {(M1, W1), (M1, W2), (M2, W1), (M2, W2), (M3, W1), (M3, W2), (W1, W2)
n (A) = 7
For event B:
B = {(M1, W1), (M1, W2), (M2, W1), (M2, W2), (M3, W1), (M3, W2)
n (B) = 6
For event c:
C = {(M1, M2), (M1, M3), (M2, M3)}
n (c) = 3
(6) One coin and one die are thrown simultaneously.
Condition for event A: To get head and an odd number.
Condition for event B: To get a head or tail and an even number.
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
Solution:
One coin and die thrown.
∴ S {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
n (s) = 12
For event A:
A = {(H, 1), (H, 3), (H, 5)}
n (A) = 3
For event B:
B = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}
n (B) = 6
For event C:
C = {Not possible}
n (C) = 0
Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 5 Probability Practice Set 5.3 Solution
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