Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Probability

Short Answer Type

991.

Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether or not the trials of drawing balls arc Bernoulli trials when after each drawn the ball drawn is (i) replaced (ii) not replaced in the urn.

85 Views

992. There is a group of 10 families each having 5 children of which 2 are boys and 3 are girls. One child is selected from each family, each child of the family being equally likely to be selected. Find the probability that among the 10 children so selected, there are exactly 5 girls.

81 Views

993. A pair of dice is thrown 4 times. If a doublet is considered a success, find the probability of 2 successes.

89 Views

994. A pair of dice is thrown 10 times. If getting a doublet is considered a success, find the probability of getting 4 successes.

120 Views

995. A box contains 100 tickets each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.

139 Views

Long Answer Type

996. A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
i) none is white? (ii) all are white? (iii) only 2 are white? (iv) at least one is white?

Here, n = 4, p = $fraction numerator 5 over denominator 5 plus 7 plus 8 end fraction space equals 1 fourth$

$straight q space equals space 1 minus straight p space equals space 1 minus 1 fourth space equals space 3 over 4$

(i) P(none is white) = P(0) = $straight C presuperscript 4 subscript 0 space open parentheses 3 over 4 close parentheses to the power of 4 minus 0 end exponent space open parentheses 1 fourth close parentheses to the power of 0 space equals space 1 cross times open parentheses 3 over 4 close parentheses to the power of 4 space cross times space 1 space equals space 81 over 256$

(ii) P(all white) = P(4) = $straight C presuperscript 4 subscript 4 space open parentheses 3 over 4 close parentheses to the power of 4 minus 4 end exponent space open parentheses 1 fourth close parentheses to the power of 4 space equals space 1 cross times 1 cross times 1 over 256 space equals 1 over 256$

(iii) P(only 2 are white) = P(2) = $straight C presuperscript 4 subscript 2 open parentheses 3 over 4 close parentheses to the power of 4 minus 2 end exponent space open parentheses 1 fourth close parentheses squared space equals space fraction numerator 4 cross times 3 over denominator 1 cross times 2 end fraction cross times 9 over 16 cross times 1 over 6 space equals 27 over 128$

(iv) P(at least one is white) = P(1) + P(2) + P(3) + P(4)

$equals space 1 minus straight P left parenthesis 0 right parenthesis space equals space 1 minus 81 over 256 space equals fraction numerator 256 minus 81 over denominator 256 end fraction space equals 175 over 256$

88 Views

997.

Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?

94 Views