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Question 14 Marks
Two coins are tossed simultaneously 300 times and the outoomes are noted as :
Two tails : 83
Ole tail :140
No tai :77
Find the probabiity of Occurrence of each of these events.
Answer
Number of times 2 tails come up $=83$
Total number of times the coins were tossed $=300$
$
\begin{aligned}
& P(2 \text { tails will come up })=\frac{\text { Number of times } 2 \text { tails come up }}{\text { Total number of times the coins were tossed }} \\
& =\frac{83}{300}
\end{aligned}
$
Number of times 1 tail come up $=140$
Total number of times the coins were tossed $=300$
$
\mathrm{P}(1 \text { tail will come up })=\frac{\text { Number of times } 1 \text { tail come up }}{\text { Total number of times the coins were tossed }}
$
$
=\frac{140}{300}=\frac{7}{15}
$
Number of times no tail come up $=77$
Total number of times the coins were tossed $=300$
$
P(\text { no tail will come up })=\frac{\text { Number of times no tail come up }}{\text { Total number of times the coins were tossed }}
$
$
=\frac{77}{300}
$
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Question 24 Marks
Two dice, one white and one red are rolled together .Find the probability of getting
(iii) a difference of 1
Answer
lf two dice are rolled then the possible outcomes are :
(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 the total number of outcomes = 36
a difference of 1
Favorable outcomes for getting a difference 1 are :
(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3 ), (4, 5), (5, 4), (5, 6), (6, 5)
So the number of possible outcomes = 10
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Question 34 Marks
Two dice, one white and one red are rolled together .Find the probability of getting
(ii) two different digits
Answer
If two dice are rolled then the possible outcomes are:
$
\begin{aligned}
& (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)
\end{aligned}
$
So the total number of outcomes $=36$
two different digits
Here we use the following formula :
$\mathrm{P}$ (Two different digits) $=1-\mathrm{P}$ (both digits are same)
Now favorable outcomes for both digits same are: $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
So the number of possible outcomes for both digits same $=6$
Thus, $\mathrm{P}$ (Two different digits) $=1-\frac{6}{36}=\frac{5}{6}$
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Question 44 Marks
Two dice, one white and one red are rolled together .Find the probability of getting
(i) a sum of 6
Answer
If two dice are rolled then the possible outcomes are:
$
\begin{aligned}
& (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)
\end{aligned}
$
So the total number of outcomes $=36$
a sum of 6
Favorable outcomes for getting a sum 6 are:
$
(1,5),(2,4),(3,3),(4,2),(5,1)
$
So number of favorable outcomes $=5$
Thus, $P($ a sum of 6$)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{5}{36}$
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[4 marks sum] - Mathematics STD 10 Questions - Vidyadip